Calculus Problem Sheet
Prof Paul Sutcliffe
1. For each of the following equations, does it represent a function y(x) ?
(a) x2 + (y − 1)2 = 1
(b) y = x2 − 2x + 1
(c) x + |y| = 1
(d) |x| + y = 1
(e) y 2 = 4x2
Solution: Applying the vertical line test yields (a) no, (b) yes, (c) no, (d) yes, (e) no.
2. State the domain and range of each of the following functions
(a) f (x) = |x − 1| − 7
√
(b) f (x) = 5 − 2x
√
(c) f (x) = 2 x2 − 3
(d) f (x) = 2x2 /(x2 + 4)
(e) f (x) = −x/(x2 − 16)
(f) f (x) = e1/(x
2 −4)
Solution: (a) Dom f = R, Ran f = [−7, ∞)
(b) Dom f = [0, ∞), Ran f = (−∞, 5]
√
√
(c) Dom f = (−∞, − 3] ∪ [ 3, ∞), Ran f = [0, ∞)
(d) Dom f = R, Ran f = [0, 2)
(e) Dom f = R\{±4}, Ran f = R
(f) Dom f = R\{±2}, Ran f = (0, e−1/4 ] ∪ (1, ∞)
3. Let f (x) = x2 + ax + 1, where a ∈ R.
Derive the allowed values of a such that ∀x ∈ R, f (x) = |f (x)|.
Solution: f = |f | iff f ≥ 0 so the required condition is equivalent to ∀x ∈ R, f (x) ≥ 0. By
completing the square f (x) = (x + 12 a)2 + 1 − 41 a2 ≥ 0 hence a2 ≤ 4 giving a ∈ [−2, 2].
4. Graph the following functions
(a) f (x) = | − x2 + 3x|
2
(b) f (x) = e−(x+1)
(c) f (x) = | − x2 − x − 1|
18
1
16
0.9
7
6
0.8
14
5
0.7
12
0.6
4
f(x)
f(x)
f(x)
10
0.5
8
3
0.4
6
0.3
4
2
0.2
1
2
Solution:
0.1
0
0
-3
-2
-1
0
1
2
x
3
4
5
6
0
-4
-3
-2
-1
0
x
1
2
3
-3
-2
-1
0
x
1
2
10
h(x)
g(x)
f(x)
9
8
7
6
5
4
3
2
1
0
-1
-2
0
Solution:
2
4
6
8
10
12
x
5. For x ∈ [0, 4π], on the same drawing, graph the following three functions
h(x) = sin x, g(x) = 1 + sin x, f (x) = 1/(1 + sin x).
6. For x ∈ [−π, π], on the same drawing, graph the following three functions
h(x) = cos x, g(x) = cos(2x), f (x) = 1/ cos(2x).
4
h(x)
g(x)
f(x)
3
2
1
0
-1
-2
-3
-4
-3
Solution:
-2
-1
0
1
2
3
x
7. For x ∈ [−2, 2], on the same drawing, graph the following three functions
h(x) = e−x , g(x) = e−x − 1, f (x) = 1/(e−x − 1)
5
h(x)
g(x)
f(x)
4
3
2
1
0
-1
-2
-3
-4
-5
Solution:
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
8. Are the following functions even, odd or neither? Justify your answers.
(a) f (x) = (x − 1)(x − 2)
P
(b) f (x) = nk=0 x2k+1
(c) f (x) = sin(x2 )
(d) f (x) =
x
(x2 +1) cos x
Solution:
(a) f (x) = x2 − 3x + 2, so f (−x) = x2 + 3x + 2. Since f (x) 6= f (−x) and f (x) 6= −f (−x)
this function is neither even nor odd.
P
P
(b) f (−x) = nk=0 (−1)2k+1 x2k+1 = − nk=0 x2k+1 = −f (x) hence this function is odd.
(c) f (−x) = sin((−x)2 ) = sin(x2 ) = f (x) hence this function is even.
(d) x is odd, but both x2 + 1 and cos x are even, hence f (x) is the product of one odd
function and two even functions and is therefore an odd function.
9. Are the following functions even, odd or neither? Justify your answers.
(a) f (x) = ex
(b) g(x) = tan x
1
(c) h(x) = xe− 2 | log x
2|
(d) k(x) = log |x|
(e) p(x) = (x3 + x)/(x3 − x)
(f) q(x) = sin2 (4x)
(g) r(x) = x2 − 4 sin x
Solution:
(a) f (−x) = e−x . Since f (x) 6= f (−x) and f (x) 6= −f (−x) this function is neither even nor
odd.
(b) g(−x) = sin(−x)/ cos(−x) = − sin x/ cos x = − tan x = −g(x) hence this function is
odd.
1
1
2
2
(c) h(−x) = −xe− 2 | log(−x) | = −xe− 2 | log x | = −h(x) hence this function is odd.
(d) k(−x) = log | − x| = log |x| = k(x) hence this function is even.
(e) x3 + x is an odd function and so is x3 − x hence p(x) is the ratio of two odd functions
and is therefore even.
(f) q(−x) = sin2 (−4x) = (sin(−4x))2 = (− sin(4x))2 = sin2 (4x) hence this function is
even.
(g) r(−x) = (−x)2 − 4 sin(−x) = x2 + 4 sin x. Since r(x) 6= r(−x) and r(x) 6= −r(−x) this
function is neither even nor odd.
10. If f : R 7→ R is an even function and g : R →
7 R is an odd function then determine
whether the following functions are even, odd or neither? Justify your answers.
(
f (x)
if x > 0
(a) f1 (x) =
−f (x) if x < 0
(b) f2 (x) = f (x) + |f (x)|
(c) f3 (x) = (g ◦ f )(x)
(d) f4 (x) = (f ◦ g)(x)
(e) f5 (x) = (g ◦ g)(x)
Solution:
(a) On R\{0}
(
(
f (−x)
if − x > 0
f (x)
if x < 0
f1 (−x) =
=
= −f1 (x) hence this function
−f (−x) if − x < 0
−f (x) if x > 0
is odd.
(b) f2 (−x) = f (−x) + |f (−x)| = f (x) + |f (x)| = f2 (x) hence this function is even.
(c) f3 (−x) = (g ◦ f )(−x) = g(f (−x)) = g(f (x)) = (g ◦ f )(x) = f3 (x) hence this function
is even.
(d) f4 (−x) = (f ◦ g)(−x) = f (g(−x)) = f (−g(x)) = f (g(x)) = (f ◦ g)(x) = f4 (x) hence
this function is even.
(e) f5 (−x) = (g ◦ g)(−x) = g(g(−x)) = g(−g(x)) = −g(g(x)) = −(g ◦ g)(x) = −f5 (x)
hence this function is odd.
11. Write each of the following functions as the sum of an even function feven (x) and an odd
function fodd (x).
(a) f (x) = x3 − 5x + 4
(b) f (x) = ex
3
(c) f (x) = log |x + 2|
(d) f (x) = x3 /(x2 − 1)
Solution:
(a) By inspection feven (x) = 4 and fodd (x) = x3 − 5x.
3
3
(b) feven (x) = 12 (f (x) + f (−x)) = 12 (ex + e−x ) and fodd (x) = 12 (f (x) − f (−x)) =
1 x3
−x3 )
2 (e − e
(c) feven (x) = 21 (f (x)+f (−x)) = 12 (log |2+x|+log |2−x|) and fodd (x) = 12 (f (x)−f (−x)) =
1
2 (log |2 + x| − log |2 − x|)
(d) Since f (−x) = −x3 /(x2 − 1) = −f (x) is an odd function feven (x) = 0 and fodd (x) =
f (x) = x3 /(x2 − 1)
12. In the following cases write a formula for the functions f ◦ g and g ◦ f and find the
domain and range of each of them.
√
(a) f (x) = x + 2 and g(x) = 1/x.
√
(b) f (x) = x2 and g(x) = 1 − x.
Solution:
q
(a) (f ◦ g)(x) = f (g(x)) = f (1/x) = x1 + 2. Dom (f ◦ g) = (−∞, − 21 ] ∪ (0, ∞)
√
Ran (f ◦ g) = [0, ∞)\{ 2}.
√
√
(g ◦ f )(x) = g(f (x)) = g( x + 2) = 1/ x + 2 Dom (g ◦ f ) = (−2, ∞), Ran (g ◦ f ) =
(0, ∞).
√
√
√
(b) (f ◦ g)(x) = f (g(x)) = f (1 − x) = (1 − x)2 = 1 − 2 x + x
Dom (f ◦ g) = [0, ∞), Ran (f ◦ g) = [0, ∞).
√
(g ◦ f )(x) = g(f (x)) = g(x2 ) = 1 − x2 = 1 − |x|
Dom (g ◦ f ) = R, Ran (g ◦ f ) = (−∞, 1].
13. Given f (x) = x − 1 and g(x) = 1/(1 + x), find
(a) (f ◦ g)( 12 )
(b) (f ◦ f )(2)
(c) (g ◦ f )(x)
(d) (g ◦ g)(2)
Solution:
(a) (f ◦ g)( 12 ) = f (2/3) = −1/3
(b) (f ◦ f )(2) = f (1) = 0
(c) (g ◦ f )(x) = g(x − 1) = 1/x
(d) (g ◦ g)(2) = g(1/3) = 3/4
14. Given u(x) = 2x − 3, v(x) = x4 and f (x) = 1/x, find
(a) (u ◦ (v ◦ f ))(x)
(b) (v ◦ (u ◦ f ))(x)
(c) (f ◦ (v ◦ u))(x)
(d) (v ◦ (f ◦ u))(x)
Solution:
(a) (u ◦ (v ◦ f ))(x) = u(v(1/x)) = u(1/x4 ) = 2/x4 − 3
(b) (v ◦ (u ◦ f ))(x) = v(u(1/x)) = v(2/x − 3) = (2/x − 3)4
(c) (f ◦ (v ◦ u))(x) = f (v(2x − 3)) = f ((2x − 3)4 ) = 1/(2x − 3)4
(d) (v ◦ (f ◦ u))(x) = v(f (2x − 3)) = v(1/(2x − 3)) = 1/(2x − 3)4
15. For each f (x) given below, find the inverse function f −1 (x) and identify its domain and
range.
(a) f (x) = x5
(b) f (x) = x3 + 1
(c) f (x) = 1/x2 , x > 0
(d) f (x) = x4 , x ≥ 0
(e) f (x) = 12 x −
7
2
(f) f (x) = 1/x3 , x 6= 0
Solution:
Write y = f −1 (x) and use f (y) = x.
1
(a) f (y) = y 5 = x hence y = x 5 = f −1 (x).
Dom f −1 = Ran f = R and Ran f −1 = Dom f = R.
1
(b) f (y) = y 3 + 1 = x hence y = (x − 1) 3 = f −1 (x).
Dom f −1 = Ran f = R and Ran f −1 = Dom f = R.
√
(c) f (y) = 1/y 2 = x hence y = 1/ x = f −1 (x).
Dom f −1 = Ran f = (0, ∞) and Ran f −1 = Dom f = (0, ∞).
1
(d) f (y) = y 4 = x hence y = x 4 = f −1 (x).
Dom f −1 = Ran f = [0, ∞) and Ran f −1 = Dom f = [0, ∞).
(e) f (y) = 12 y − 72 = x hence y = 2x + 7 = f −1 (x).
Dom f −1 = Ran f = R and Ran f −1 = Dom f = R.
1
(f) f (y) = 1/y 3 = x hence y = x− 3 = f −1 (x).
Dom f −1 = Ran f = R\{0} and Ran f −1 = Dom f = R\{0}.
16. Which of the following functions are injective? Find the inverses of those which are and
specify the domain of the inverse.
(a) f (x) = (1 + 3x)3 on R
(b) f (x) = (1 − x)2 on [1, 2]
(c) f (x) = (1 − x)2 on [−1, 2]
(d) f (x) = (x − 1)/(x + 2) on R\{−2}
(e) f (x) = x2 + 2x − 1 on [−1, 1]
(f) f (x) = x2 + 2x − 1 on [−2, 2]
Solution:
(a). It is injective. Apply horizontal line test or
f (x1 ) = f (x2 ) iff (1 + 3x1 )3 = (1 + 3x2 )3 iff (1 + 3x1 ) = (1 + 3x2 ) iff x1 = x2 .
1
Write y = f −1 (x) and use f (y) = x. So f (y) = (1+3y)3 = x hence y = 31 (x 3 −1) = f −1 (x).
Dom f −1 = Ran f = R.
(b). It is injective. Apply horizontal line test.
√
Write y = f −1 (x) and use f (y) = x. So f (y) = (y − 1)2 = x hence y = 1 + x = f −1 (x).
Dom f −1 = Ran f = [0, 1].
(c). It is not injective. Apply horizontal line test or eg. f (2) = 1 = f (0).
(d). It is injective. Apply horizontal line test.
Write y = f −1 (x) and use f (y) = x.
So f (y) = (y − 1)/(y + 2) = x hence y = (2x + 1)/(1 − x) = f −1 (x).
Dom f −1 = Ran f = R\{1}.
(e) It is injective. Apply horizontal line test or f (x) = x2 + 2x − 1 = (x + 1)2 − 2 so [−1, 1]
is only on one side of the turning point x = −1 of√
the quadratic. Write y = f −1 (x) and use
f (y) = x. So f (y) = (y + 1)2 − 2 = x hence y = x + 2 − 1 = f −1 (x).
Dom f −1 = Ran f = [−2, 2].
(f) It is not injective. Apply horizontal line test or eg. f (−2) = −1 = f (0).
17. Complete the following tables.
g(x)
x−7
x+2
x
x−1
f (x)
(f ◦ g)(x)
√
x
3x
√
√
x−5
x2 − 5
x
x−1
1+
Solution:
1
x
g(x)
x−7
x+2
x2
f (x)
√
x
3x
√
x−5
x
x−1
1
x−1
x
x−1
1 + x1
x
(f√◦ g)(x)
x−7
3x
√ +6
x2 − 5
x
x
g(x) f (x) (f ◦ g)(x)
1/x
x
1
|x|
x−1
x−1
√
√x
x
x
g(x)
1/x
f (x)
1
x−1
|x|
x+1
x2
√
x
x−1
√x
1
x
x
x2
x
x+1
|x|
|x|
(f ◦ g)(x)
x
1
|x−1|
x
x+1
|x|
|x|
18. For x 6= 0, 1 define the following six functions
1
1
x−1
x
, f3 (x) = 1 − x, f4 (x) =
, f5 (x) =
, f6 (x) =
.
x
1−x
x
x−1
f1 (x) = x, f2 (x) =
These have the property that the composition of any two of these functions is again one
of these functions. Complete the following table
◦
f1
f2
f3
f4
f5
f6
f1
f2
f3
f4
f5
f6
f3
f3
f4
f1
f2
f6
f5
f4
f4
f3
f6
f5
f1
f2
f5
f5
f6
f2
f1
f4
f3
f4
Solution:
◦
f1
f2
f3
f4
f5
f6
f1
f1
f2
f3
f4
f5
f6
f2
f2
f1
f5
f6
f3
f4
f6
f6
f5
f4
f3
f2
f1
19. State any vertical and horizontal asymptotes of the following functions
(a) f (x) = −2x2 /(x2 − 9)
(b) f (x) = 3x/(x2 + 1)
(c) f (x) = x3 /(x2 + 2)
(d) f (x) =
3x5 +5x2
(x−1)(7x4 +9)
Solution:
(a) Vertical asymptotes x = ±3, horizontal asymptote y = −2.
(b) Horizontal asymptote y = 0.
(c) No asymptotes.
(d) Vertical asymptote x = 1, horizontal asymptote y = 3/7.
20. Write each of the following rational functions as the sum of a polynomial and a proper
rational function
(a)
x6 +x4 −x3 +2x2 +5
,
x2 +1
(b)
x6 +2x4 +3x2 +5
,
x4 +2
Solution:
6
4
3 +2x2 +5
(a) x +x −x
= x4 − x + 2 +
x2 +1
(b)
(c)
(d)
2 +1
x6 +2x4 +3x2 +5
= x2 + 2 + xx4 +2
x4 +2
2x5 +4x3 −1
x2 +1
= x2 + 2 + 2x
3 −1
2x3 −1
x7 +6x4 +8x+1
4
= x + 2x + x31+4
x3 +4
(c)
x+3
x2 +1
2x5 +4x3 −1
,
2x3 −1
(d)
x7 +6x4 +8x+1
.
x3 +4
21. Evaluate each of the following expressions (without using a calculator)
(a) log e
(b) log3
1
9
(c) log 1 4
16
(d) log8 8−3
(e) log5 625
(f) log10 10n ,
(g)
n∈Z
log(ne)
,
m log n+log em
Solution:
(a) 1.
n, m > 0
(b) −2.
(c) − 21 .
(d) −3.
(e) 4.
(f) n.
(g)
log(ne)
m log n+log em
=
log n+1
m log n+m
22. Demonstrate graphically that ||x| − 1| ≤ |x − 1|, ∀ x ∈ R. Prove this inequality (the
graphical demonstration should provide a hint).
Solution: Define f (x) = |x − 1| − ||x| − 1|. We need to show that f (x) ≥ 0, ∀ x ∈ R.
Figure 1: The red curve is the graph of ||x| − 1|. This is obtained by shifting the graph of |x|
vertically down by 1 unit and then reflecting in the x-axis any portion of the curve below this
axis. The green curve is the graph of |x − 1|. This is obtained by shifting the graph of |x| to
the right by 1 unit. The inequality corresponds to the fact that the red curve is never above
the green curve.
The graph suggests we consider three regions.
If x ≥ 0 then f (x) = |x − 1| − |x − 1| = 0 ≥ 0.
If −1 ≤ x ≤ 0 then f (x) = −(x − 1) − | − x − 1| = −x + 1 − (x + 1) = −2x ≥ 0.
If x ≤ −1 then f (x) = −(x − 1) − | − x − 1| = −x + 1 − (−x − 1) = 2 ≥ 0.
Together these three regions cover R hence we have shown that f (x) ≥ 0, ∀ x ∈ R.
23. Consider the given graph of the function f (x). Are the following statements true or false?
(a) limx→0 f (x) exists,
(b) limx→0 f (x) = 0,
(d) limx→1 f (x) = 1,
(e) limx→1 f (x) = 0,
Solution:
(a) true,
(c) false,
(b) true,
(d) false,
(c) limx→0 f (x) = 1
(f) limx→a f (x) exists ∀ a ∈ (−1, 1).
(e) false,
(f) true.
=
1
m.
2
1.5
1
y
0.5
0
-0.5
-1
-1.5
-2
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
1
0.5
y
0
-0.5
-1
-1.5
-2
-1
-0.5
0
0.5
1
x
1.5
2
2.5
3
24. Consider the given graph of the function f (x). Are the following statements true or false?
(a) limx→2 f (x) does not exist,
(b) limx→2 f (x) = 1,
(d) limx→a f (x) exists ∀ a ∈ (−1, 1)
Solution:
(a) false,
(b) true,
(c) true,
(c) limx→1 f (x) does not exist,
(e) limx→a f (x) exists ∀ a ∈ (1, 3).
(d) true,
(e) true.
25. If f (x) > 0 ∀ x 6= a and limx→a f (x) = L, can we conclude that L > 0? Justify your
answer.
Solution:
No. An example is provided by f (x) = x2 with a = 0 so that L = 0 which is not positive.
26. Justify whether the following statement is true or false.
p
If limx→a f (x) exists then so does limx→a f (x).
Solution: False. An example is provided by f (x) = −1, with a = 0.
p
Here limx→0 f (x) exists (and is equal to −1) but f (x) is not a real function.
27. Calculate the following limits
(a) limx→0 (2 − x),
(b) limx→−1
Solution: (a) 2, (b) −1,
3x2
,
2x−1
(c) π/2,
(d)
(c) limx→π/2 x sin x,
(d) limx→π
1
π−1 .
28. Calculate the following limits
(a) limx→1
x2 +x−2
,
x2 −x
(b) limx→1
x4 −1
,
x3 −1
(e) limx→1
√ x−1 ,
x+3−2
(f) limx→4
2
4x−x
√ .
2− x
cos x
.
1−π
(c) limx→2
x3 −8
,
x4 −16
√
(d) limx→9
x−3
,
x−9
Solution:
(a)limx→1
x2 +x−2
x2 −x
(x+2)(x−1)
= limx→1 x+2
x = 3.
x(x−1)
2
2
4
+1)(x+1)(x−1)
−1
= limx→1 (x(x−1)(x
= limx→1 (x x+1)(x+1)
= 4/3.
(b) limx→1 xx3 −1
2 +x+1)
2 +x+1
2
3 −8
(x−2)(x +2x+4)
x2 +2x+4
(c) limx→2 xx4 −16
= limx→2 (x+2)(x−2)(x
2 +4) = limx→2 (x+2)(x2 +4) = 3/8.
√
√
√
x−3)( x+3)
x−3
x−9
1
√
√
= limx→9 ((x−9)(
(d) limx→9 x−9
= limx→9 (x−9)(
= limx→9 √x+3
x+3)
x+3)
√
√
x+3+2)
x+3+2)
x−1
√
= limx→1 (√(x−1)(
= limx→1 (x−1)(
= 4.
(e) limx→1 √x+3−2
x+3−4
x+3−2)( x+3+2)
√
√
2
x(4−x)(2+ x)
√ = limx→4
√
√
= limx→4 (x(2 + x)) = 16.
(f) limx→4 4x−x
2− x
(2− x)(2+ x)
= limx→1
= 1/6.
29. Calculate the limit as x → 0 of the following
(a)
1−cos 2x
,
x
(b)
Solution:
(a) limx→0
(b)
1−cos x
,
x2
1−cos 2x
x
x
limx→0 1−cos
x2
(c)
tan 2x
,
x
= limx→0
=
(d)
2(1−cos 2x)
2x
x2
,
1−cos 2x
= limu→0
x)(1+cos x)
limx→0 (1−cos
x2 (1+cos x)
x2
.
1−cos 4x
2(1−cos u)
u
= limx→0
sin x
x
1
1+cos x
= 1/2.
sin x 2 cos x
( x )( cos 2x ) = 2.
tan 2x
x
(d) limx→0
x2
1−cos 2x
= limx→0
= limx→0
= 1/2.
(e) limx→0
x2
1−cos 4x
=
=
= 1/8.
30. Does limx→0
sin(x+|x|)
x
= limx→0
2 sin x cos x
x cos 2x
= 0.
2
(c) limx→0
= limx→0
sin 2x
x cos 2x
(e)
x2 (1+cos 2x)
(1−cos 2x)(1+cos 2x)
x2 (1+cos 4x)
limx→0 (1−cos
4x)(1+cos 4x)
= limx→0
(2x)2 (1+cos 2x)
4 sin2 2x
(4x)2 (1+cos 4x)
limx→0 16 sin2 4x
exist? If so, find it.
2x
Solution: For x > 0, sin(x+|x|)
= sinx2x . Hence limx→0+ sin(x+|x|)
= limx→0+ sinx2x = limx→0+ 2 sin
= 2.
x
x
2x
sin(x+|x|)
sin(x+|x|)
For x < 0,
= 0. Hence limx→0−
= 0.
x
x
The left-sided and right-sided limits exist but are not equal, hence the limit does not exist.
31. Calculate limx→π/2 {(x − π/2) tan x}.
Solution:
Set u = x−π/2 then limx→π/2 (x−π/2) tan x = limu→0 u tan(u+π/2) = limu→0
u
= limu→0 −usincos
u = −1.
u sin(u+π/2)
cos(u+π/2)
32. In each case either evaluate the limit or state that no limit exists
(a) limx→3
x2 +x+12
,
x−3
(e) limh→0
1−1/h2
,
1+1/h2
(b) limx→3
(f) limh→0
x2 +x−12
,
x−3
(c) limx→3
(x2 +x−12)2
,
x−3
(d) limx→3
1+1/h
.
1+1/h2
Solution:
(a) no limit exists,
x2 +x−12
= limx→3 (x+4)(x−3)
= 7.
x−3
x−3
(x2 +x−12)2
(x+4)2 (x−3)2
(c) limx→3
= limx→3
= limx→3 (x
x−3
x−3
(x2 +x−12)
(x+4)(x−3)
x+4
(d) (x−3)2 = (x−3)2 = x−3 hence no limit exists.
2
2
(e) limh→0 1−1/h
= limh→0 hh2 −1
= −1.
1+1/h2
+1
2
1+1/h
h +h
(f) limh→0 1+1/h
2 = limh→0 h2 +1 = 0.
(b) limx→3
+ 4)2 (x − 3) = 0.
(x2 +x−12)
,
(x−3)2
33. Calculate the limit as x → ∞ of the following
(a)
6x+7
,
1−2x
(b)
(x2 +1)2 −(x2 −1)2
,
(x+2)3 −(x+1)3
(c)
x2
.
x2 +sin2 x
Solution:
(a) limx→∞
6x+7
1−2x
(b) limx→∞
(x2 +1)2 −(x2 −1)2
(x+2)3 −(x+1)3
= limx→∞
(c) First note that 0 ≤
As limx→∞
Thus
1
x2
sin2 x
x2
6+ x7
1
−2
x
= −3.
4x2
3x2 +9x+7
= limx→∞
= limx→∞
7
x2
= 4/3.
1
.
x2
≤
= 0 then by the pinching theorem limx→∞
x2
limx→∞ x2 +sin
2x
4
3+ x9 +
= limx→∞
1
2
1+ sin 2 x
sin2 x
x2
= 0.
= 1.
x
34. Calculate the following limits
(a) limx→∞ x sin
1
,
x
(b)
limx→∞ cos(1/x)
,
1+(1/x)
(d) limx→∞ (3 + x2 ) cos(1/x),
1/x
1
x
(c) limx→∞
,
(e) limx→∞ {( x32 − cos(1/x))(1 + sin(1/x))}.
Solution:
Set u = 1/x in each case
(a) limx→∞ x sin x1 = limu→0
cos(1/x)
1+(1/x) =
1/x
limx→∞ x1
=
(b) limx→∞
(c)
sin u
u = 1.
u
limu→0 cos
1+u = 1.
limu→0 uu = limu→0 exp(log uu ) = limu→0 exp(u log u) = e0 = 1.
(d) limx→∞ (3 + x2 ) cos(1/x) = limu→0 (3 + 2u) cos u = 3.
(e) limx→∞ ( x32 − cos(1/x))(1 + sin(1/x)) = limu→0 (3u2 − cos u)(1 + sin u) = −1.
35. For each of the following statements, either give a proof that it is true or a counter
example to show that it is false:
(a) If g(x) > 0 ∀ x > 0 and limx→∞ (f (x) − g(x)) = 0 then limx→∞ (f (x)/g(x)) = 1.
(b) If g(x) > 0 ∀ x > 0 and limx→∞ (f (x)/g(x)) = 1 then limx→∞ (f (x) − g(x)) = 0.
Solution:
(a) False. A counter example is provided by f (x) = 1/x and g(x) = 2/x.
(b) False. A counter example is provided by f (x) = x and g(x) = x + 1.
36. In each case either evaluate the limit or state that no limit exists
(a) limu→−5
u2
,
5−u
(e) limx→−2
√ x+2
,
x2 +5−3
(f) limx→∞
(i) limx→−3
x+3
,
x2 +4x+3
(j) limx→2
(b) limy→0 (2y − 8)1/3 ,
(c) limx→0
−3x4 +x2 +1
,
−5x4 −1
√
x2 +12−4
,
x−2
(x−2)(1−cos 3x)
,
2x
(g) limt→0
(k) limt→1
Solution:
(a) 5/2, (b) −2, (c) 0, (d) 1/10, (e) −3/2,
(i) −1/2, (j) 1/2, (k) 3/2, (l) no limit exists.
5t3 +8t2
,
3t2 −16t4
t2 +t−2
,
t2 −1
(f) 3/5,
(d) limt→5
(h) limx→3
(l) limt→−∞
(g) 8/3,
t−5
,
t2 −25
tan(2(x−3))
,
x−3
t3 +1
.
t2 +1
(h) 2,
37. Sketch the graph of the function f (x) and classify any discontinuities, where


2x − 1 if x < 1
f (x) = 0
if x = 1


2
1/x
if x > 1
Solution: Removable discontinuity at x = 1.
1
0.5
y
0
-0.5
-1
0
0.5
1
1.5
x
2
2.5
3
38. Sketch the graph of the function f (x) and classify any discontinuities, where
(
(x − 3)/(x2 − 9) if x 6= ±3
f (x) =
1/6
if x = ±3
Solution: Infinite discontinuity at x = −3.
4
3
2
y
1
0
-1
-2
-3
-4
-6
-4
-2
0
2
4
x
39. Sketch the graph of the function f (x) and


−1
f (x) = x3


1
classify any discontinuities, where
if x < −1
if − 1 ≤ x ≤ 1
if x > 1
Solution: No discontinuities.
1
y
0.5
0
-0.5
-1
-2
-1.5
-1
-0.5
0
x
0.5
1
1.5
2
40. Use the limit definition of the derivative to calculate the derivative of the following
functions
√
(a) f (x) = sin x, (b) f (x) = x x, (c) f (x) = cos2 x.
Solution:
x
x sin h−sin x
(a) f 0 (x) = limh→0 sin(x+h)−sin
= limh→0 sin x cos h+cos
= limh→0
h
h
1−cos h
sin h
= − sin x limh→0 h + cos x limh→0 h = 0 sin x + 1 cos x = cos x.
(b) f 0 (x) = limh→0
√
√
(x+h) x+h−x x
h
√
(x+h) x+h−x x
= limh→0
h
= limh→0 h
(x+h)3 −x3
√
√
(x+h) x+h+x x
√
= limh→0
3x2√
+3xh+h2√
(x+h) x+h+x x
sin x cos h+cos x sin h−sin x
h
√
√
(x+h) x+h+x x
√
√
(x+h) x+h+x x
=
3x√2
2x x
=
3√
2 x.
2
2
cos2 (x+h)−cos2 x
= limh→0 (cos x cos h−sinhx sin h) −cos x
h
2
2
x sin h cos h+sin2 x sin2 h
limh→0 cos x(cos h−1)−2 sin x cos
h
2
2
2
2
2
sin x cos x sin h cos h
sin x cos x sin(2h)
limh→0 (sin x−cos x) sin h−2
= limh→0 (sin x−cos x)(1−cos(2h))−2
h
2h
0(sin2 x − cos2 x) + 1(−2 sin x cos x) = −2 sin x cos x.
(c) f 0 (x) = limh→0
=
=
=
41. Find the Cartesian equation for the tangent to the graph of f : R 7→ R : x 7→ 5x2 − 4x
at the point (1, 1).
Solution: f (x) = 5x2 − 4x, hence f 0 (x) = 10x − 4, with f 0 (1) = 6. The Cartesian equation for
the tangent at (1, 1) is y = f (1) + f 0 (1)(x − 1) = 1 + 6(x − 1) = 6x − 5.
42. Find the slope of the straight line√which passes through the point (−2, 0) and is also
tangential to the graph of f (x) = x at some point.
√
Solution: Let x = a > 0 be the point at which the line is tangential to the graph of f (x) = x.
√
As f 0 (x) = 1/(2 x) then the tangent line is given by
√
1
y = f (a) + f 0 (a)(x − a) = a + 2√
(x − a). For (−2, 0) to be on this point requires that
a
√
√
1
0 = a + 2√a (−2 − a), with solution a = 2. The slope is then f 0 (2) = 1/(2 2).
43. Show that if g(x) is continuous at x = 0 then g(x) tan x is differentiable at x = 0.
Solution: Let f (x) = g(x) tan x then we need to show that limh→0
(0)
tan 0
limh→0 f (h)−f
= limh→0 g(h) tan h−g(0)
h
h
=
limh→0 g(h)
limh→0
1
cos h
limh→0
=
f (h)−f (0)
h
exists.
limh→0 g(h)htan h
sin h
h
= g(0)(1)(1) = g(0)
where we have made use of the continuity of g(x) and 1/ cos x at x = 0.
Hence f (x) = g(x) tan x is differentiable at x = 0 with f 0 (0) = g(0).
44. Calculate the following derivatives
p
d
d
(a) dy
y 2 + y, (b) da
(ax2 + bx + c),
(d)
d x
a ,
dx
where a > 0,
(e)
d x
a ,
da
(c)
d
dt
t2 −1
t2 +1
,
where x is a rational number.
Solution:
p
2y+1
y2 + y = √
,
2
(a)
d
dy
(d)
d x
dx a
2
=
d log ax
dx e
y +y
=
(b)
d x log a
dx e
d
2
da (ax
+ bx + c) =
x2 ,
= log a ex log a = log a ax ,
t2 −1
t2 +1
(c)
d
dt
(e)
d x
da a
=
4t
.
(t2 +1)2
= xax−1 .
45. Let f : R 7→ R be a differentiable function that satisfies 2f (x) + ex
and has a continuous derivative. Find f (0) and f 0 (0).
2 f (x)
− sin f (x) = 1
Solution:
Evaluating the given equation at x = 0 yields 2f (0) + 1 − sin f (0) = 1, that is
2f (0) = sin f (0). The only solution of this equation is f (0) = 0.
Differentiating the given equation with respect to x gives
2
2f 0 (x) + (2xf (x) + x2 f 0 (x))ex f (x) − f 0 (x) cos x = 0, and after setting x = 0 this becomes
2f 0 (0) − f 0 (0) = 0, that is, f 0 (0) = 0.
46. Explicitly write out the Leibniz rule for
derivative of x4 cos x.
Solution:
d4
(f (x)g(x))
dx4
and use this to calculate the fourth
d4
(f (x)g(x)) = f (4) (x)g(x)+4f 000 (x)g 0 (x)+6f 00 (x)g 00 (x)+4f 0 (x)g 000 (x)+f (x)g (4) (x).
dx4
= x4 , f 0 (x) = 4x3 , f 00 (x) = 12x2 , f 000 (x) = 24x, f (4) (x) = 24.
f (x)
g(x) = cos x, g 0 (x) = − sin x, g 00 (x) = − cos x, g 000 (x) = sin x, g (4) (x) = cos x.
d4
(x4 cos x)
dx4
= (24 − 72x2 + x4 ) cos x + (−96x + 16x3 ) sin x.
47. Given f (x) = 4x + 3 and g(x) = 1/(4 + x2 )2 , find (f ◦ g)(x) and (g ◦ f )(x).
Calculate (f ◦ g)0 (0) and (g ◦ f )0 (0).
Solution:
(f ◦ g)(x) = f (1/(4 + x2 )2 ) =
f 0 (x) = 4 and g 0 (x) =
g)0 (0) = 0.
4
(4+x2 )2
−4x
,
(4+x2 )3
(g ◦ f )0 (x) = g 0 (f (x))f 0 (x) =
+ 3,
hence (f ◦ g)0 (x) = f 0 (g(x))g 0 (x) =
−16(4x+3)
,
(4+(4x+3)2 )3
1
.
(4+(4x+3)2 )2
−16x
, giving (f
(4+x2 )3
(g ◦ f )(x) = g(4x + 3) =
giving (g ◦ f )0 (0) = −48/133 = −48/2197.
48. Use L’Hopital’s rule to calculate the limit as x → 0 of the following
(a)
1−cos 2x
,
x
(b)
1−cos x
,
x2
(c)
tan 2x
,
x
(d)
x2
,
1−cos 2x
(e)
x2
.
1−cos 4x
Solution:
(a) f (x) = 1 − cos(2x), g(x) = x, are differentiable and satisfy f (0) = g(0) = 0.
f 0 (x) = 2 sin(2x), f 0 (0) = 0, g 0 (x) = 1 6= 0.
limx→0 f (x)/g(x) = limx→0 f 0 (x)/g 0 (x) = f 0 (0)/g 0 (0) = 0/1 = 0.
(b) f (x) = 1 − cos x, g(x) = x2 , are twice differentiable and satisfy f (0) = g(0) = 0.
f 0 (x) = sin x, f 0 (0) = 0, g 0 (x) = 2x, g 0 (0) = 0.
f 00 (x) = cos x, f 00 (0) = 1, g 00 (x) = 2 6= 0.
limx→0 f (x)/g(x) = limx→0 f 0 (x)/g 0 (x) = limx→0 f 00 (x)/g 00 (x) = f 00 (0)/g 00 (0) = 1/2.
(c) f (x) = tan(2x), g(x) = x, are differentiable and satisfy f (0) = g(0) = 0.
f 0 (x) = 2 sec2 (2x), f 0 (0) = 2, g 0 (x) = 1 6= 0.
limx→0 f (x)/g(x) = limx→0 f 0 (x)/g 0 (x) = f 0 (0)/g 0 (0) = 2/1 = 2.
(d) f (x) = x2 , g(x) = 1 − cos(2x), are twice differentiable and satisfy f (0) = g(0) = 0.
f 0 (x) = 2x, f 0 (0) = 0, g 0 (x) = 2 sin(2x), g 0 (0) = 0.
f 00 (x) = 2, g 00 (x) = 4 cos(2x) 6= 0 for x sufficiently close to x = 0. Also, g 00 (0) = 4.
limx→0 f (x)/g(x) = limx→0 f 0 (x)/g 0 (x) = limx→0 f 00 (x)/g 00 (x) = f 00 (0)/g 00 (0) = 42 = 12 .
(e) f (x) = x2 , g(x) = 1 − cos(4x), are twice differentiable and satisfy f (0) = g(0) = 0.
f 0 (x) = 2x, f 0 (0) = 0, g 0 (x) = 4 sin(4x), g 0 (0) = 0.
f 00 (x) = 2, g 00 (x) = 16 cos(4x) 6= 0 for x sufficiently close to x = 0. Also, g 00 (0) = 16.
limx→0 f (x)/g(x) = limx→0 f 0 (x)/g 0 (x) = limx→0 f 00 (x)/g 00 (x) = f 00 (0)/g 00 (0) = 81 .
◦
49. Find the radius of a sphere at which the rate of change of its surface area with respect
to its radius is 2 cm.
Solution: Use length units of cm and area units of cm2 . If the sphere has radius r then its area
is A = 4πr2 . The rate of change is dA
dr = 8πr. This is equal to 2 when r = 1/(4π) = 0.08
(to 2 decimal places). Hence the required radius is 0.08cm.
50. Find an expression for
(a) 2y
3/2
dy
dx
(b) xy 2 − 4x3/2 − y = 0,
+ xy − x = 0,
2
(d) (3xy + 7) = 6y,
in the following cases
(e) x + tan(xy) = 0,
(c) x + sin y = xy,
(f) cosh x + sinh(xy) = 0.
Solution:
√
√
(a) 3 yy 0 + y + xy 0 − 1 = 0 hence y 0 = (1 − y)/(3 y + x).
√
√
(b) y 2 + 2xyy 0 − 6 x − y 0 = 0 hence y 0 = (6 x − y 2 )/(2xy − 1).
(c) 1 + y 0 cos y = y + xy 0 hence y 0 = (y − 1)/(cos y − x).
(d) 2(3xy + 7)(3y + 3xy 0 ) = 6y 0 hence y 0 = y(3xy + 7)/(1 − 7x − 3x2 y).
(e) 1 + sec2 (xy)(y + xy 0 ) = 0 hence y 0 = −(cos2 (xy) + y)/x.
(f) sinh x + cosh(xy)(y + xy 0 ) = 0 hence y 0 = − x1 (y + sinh x/ cosh(xy)).
51. Assume that the following equations each define y as a differentiable function of x.
Calculate
dy
dx
2
at the given point
(a) x3 − 2y + xy = 0,
(1, 1).
2
(b) xy + y − 3x − 3 = 0,
(c) xey + sin(xy) + y = log 2,
(−1, 1).
(0, log 2).
Solution:
(a) 3x2 − 4yy 0 + y + xy 0 = 0 at (x, y) = (1, 1) this becomes 3 − 4y 0 + 1 + y 0 = 0 so y 0 = 4/3.
(b) y + xy 0 + 2yy 0 − 3 = 0 at (x, y) = (−1, 1) this becomes 1 − y 0 + 2y 0 − 3 = 0 so y 0 = 2.
(c) ey +xy 0 ey +(y +xy 0 ) cos(xy)+y 0 = 0 at (x, y) = (0, log 2) this becomes 2+log 2+y 0 = 0
so y 0 = −2 − log 2.
52. Find all local maxima and minima, and hence the global extreme values, of x4 − 2x2 + 1
in the interval [−2, 2].
Solution: Note that f (x) = x4 − 2x2 + 1 is twice differentiable at all points.
f 0 (x) = 4x(x2 − 1), hence f 0 (x) = 0 at x = 0, ±1, which are all in [−2, 2].
Also, f 00 (x) = 4(3x2 − 1).
f 00 (0) = −4 < 0, hence x = 0 is a local maximum with f (0) = 1.
f 00 (±1) = 8 > 0, hence x = ±1 are local minima with f (±1) = 0.
f 0 (−2) = −24 < 0, hence x = −2 is an endpoint maximum with f (−2) = 9.
f 0 (2) = 24 > 0, hence x = 2 is an endpoint maximum with f (2) = 9.
The global minimum is therefore 0 and the global maximum is 9.
53. Find the global extreme values of f (x) = x|x2 − 6| − 23 x2 + 2 in [−2, 4].
√
Solution: Note that f (x) = x|x2 − 6| − 32 x2 + 2 is differentiable everywhere except at x = ± 6,
√
although x = − 6 is outside the interval [−2, 4].
√
First restrict to x ∈ (−2, 6), then f (x) = −x(x2 − 6) − 23 x2 + 2 with
√
f 0 (x) = −3x2 + 6 − 3x = −3(x + 2)(x − 1). Hence f 0 (x) = 0 in (−2, 6) iff x = 1.
f (1) = 11
2 .
√
Now restrict to x ∈ ( 6, 4), then f (x) = x(x2 − 6) − 32 x2 + 2 with
√
f 0 (x) = 3x2 − 6 − 3x = 3(x − 2)(x + 1). Hence f 0 (x) 6= 0 in ( 6, 4).
√
The only other possibilities for the global extreme √
values are x = 6 (where f (x) is not
differentiable) and the endpoints x = −2, 4. Now f ( 6) = −7, f (−2) = −8, f (4) = 18.
The global maximum is max{ 11
2 , −7, −8, 18} = 18.
11
The global minimum is min{ 2 , −7, −8, 18} = −8.
54. Find the global extreme values of f (x) = 13 x3 − 3x + |x2 − 4| in [−2, 4].
Solution: Note that f (x) =
x3
3
x3
3
− 3x + |x2 − 4| is differentiable everywhere except at x = ±2.
− 3x + 4 − x2 with
For x ∈ (−2, 2), f (x) =
f 0 (x) = x2 − 3 − 2x = (x − 3)(x + 1) = 0 iff x = −1.
f (−1) =
17
3 .
3
For x ∈ (2, 4), then f (x) = x3 − 3x + x2 − 4 with
f 0 (x) = x2 − 3 + 2x = (x + 3)(x − 1) 6= 0.
10
64
Now f (2) = − 10
3 , f (−2) = 3 , f (4) = 3 .
10
The global maximum is 64
3 and the global minimum is − 3 .
55. Find the global maximum (if it exists) of each of the following
(a) f (x) = x4 − 2x2 in [ 13 , 43 ],
(b) f (x) = x4 − 2x2 in [− 31 , 43 ],
(c) f (x) = x4 − 2x2 in [− 13 , 2],
√
(e) f (x) = 1 − |1 − x2 | in [0, 2],
(d) f (x) = x4 − 2x2 in (0, 1],
(f) f (x) = x/(x2 + 1) in x ≥ 0,
(g) f (x) = x cos( x1 )/(x + 1) in x ≥ 1.
Solution:
(a) f 0 (x) = 4x(x2 − 1) = 0 in ( 13 , 43 ) iff x = 1.
4
32
17
f (1) = −1, f ( 31 ) = − 17
81 , f ( 3 ) = − 81 . Global maximum is − 81 .
1 4
0
2
(b) f (x) = 4x(x − 1) = 0 in (− 3 , 3 ) iff x = 0, 1.
4
32
f (0) = 0, f (1) = −1, f (− 31 ) = − 17
81 , f ( 3 ) = − 81 . Global maximum is 0.
(c) f 0 (x) = 4x(x2 − 1) = 0 in (− 13 , 2) iff x = 0, 1.
f (0) = 0, f (1) = −1, f (− 31 ) = − 17
81 , f (2) = 8. Global maximum is 8.
0
2
(d) f (x) = 4x(x − 1) < 0 in (0, 1) so f (x) decreases in this interval. As the left-hand point
is not contained in the given interval (0, 1] then there is no global maximum in (0, 1].
√
(e) f (x) = 1 − |1 − x2 | is differentiable in (0, 2) except at x = 1.
For x ∈ (0, 1), f (x) = 1 − (1 − x2 ) = x2 , so f 0 (x) = 2x 6= 0.
√
For x ∈ (1, 2), f (x) = 1 + (1 − x2 ) = 2 − x2 , so f 0 (x) = −2x 6= 0.
√
f (1) = 1, f (0) = 0, f ( 2) = 0, hence the global maximum is 1.
(f) For x > 0, f 0 (x) = (1 − x2 )/(1 + x2 )2 = 0 iff x = 1.
f (1) = 12 , f (0) = 0, limx→∞ f (x) = 0, hence the global maximum is 21 .
(g) For x ≥ 1, f 0 (x) =
sin( x1 )
x(1+x) > 0, thus f (x) is increasing for x ≥ 1.
u
limx→∞ {x cos( x1 )/(x + 1)} = limu→0 cos
1+u = cos 0 = 1.
cos( x1 )
(1+x)2
+
In fact limx→∞ f (x) =
There is no global maximum in x ≥ 1.
56. A group of Chilean miners are trapped underground at a depth of 300 metres. A rescue
team starts at the bottom of an abandoned mine shaft that is 600 metres West of the
trapped miners and has a depth of 100 metres. The rescue team must dig a tunnel to
the trapped miners that has an initial horizontal segment followed by a segment directly
towards the trapped miners. At a depth of 100 metres the rock is soft and it takes only 5
minutes to dig one horizontal metre. However, at any depth below this, the rock is hard
and it takes 13 minutes to dig a distance of one metre. Calculate the minimal number
of hours it takes to tunnel to the trapped miners.
Solution: Use length units of metres and time units of minutes. Let the horizontal tunnel have a
length 600 − x, wherep
x ∈ [0, 600]. Then the distance from the end of the horizontal
p tunnel to
the trapped miners is x2 + (200)2 . The time taken is T (x) = 5(600−x)+13 x2 + (200)2 .
dT
dT
2
2
6
2
√ 13x
dx = −5 + x2 +(200)2 , therefore dx = 0 iff 25(x + 40000) = 169x , ie. 10 = 144x giving
p
2
2
x = 1000/12 = 250/3. At this value T (250/3) = 53 (1800 − 250) + 130
3 (5) 5 + (12) =
50
50
5
3 (1550 + (130)13) = 3 (155 + 169) = 3 (324) = 5400.
√
Check the endpoints: T (0) = 5600 and T (600) = 2600 10 > 5400.
Thus the minimal time is T = 5400 minutes ie. 5400/60 = 90 hours.
57. Let F (x) =
Rx
t sin t dt. Calculate F (π), F 0 (x) and F 0 (π/2).
π
Solution: F (π) = 0, F 0 (x) = x sin x and F 0 (π/2) = π/2.
58. Let
x2
Z
F (x) = −
0
2
dt.
3 + et
Find all critical points of F (x) and determine whether they are local minima, maxima
or points of inflection. Prove that F (300) > F (310).
2
Solution: F 0 (x) = −4x/(3 + ex ) hence F 0 (x) = 0 iff x = 0.
For x < 0, F 0 (x) > 0 whereas for x > 0, F 0 (x) < 0, hence x = 0 is a local maximum.
As F 0 (x) < 0 for x > 0 then F (x) is strictly monotonic decreasing in (0, ∞)
hence F (310) < F (300).
59. Calculate the derivative of F (x) where
x3
Z
(a) F (x) =
1
Z
t cos t dt,
(t − sin2 t) dt,
(b) F (x) =
x2
0
Z
cos x
(c) F (x) =
√
√
1 − t2 dt,
Z
(d) F (x) =
0
1
Solution: (a) F 0 (x) = 3x5 cos(x3 ),
(c) F 0 (x) = − sin x | sin x|,
x
t2
dt
1 + t4
(b) F 0 (x) = −2x(x2 − sin2 (x2 )),
(d) F 0 (x) =
√
1
x2
√
2 x 1+x2
√
=
x
.
2(1+x2 )
60. Calculate the derivative of the following functions
Z
t4
(a) G(t) =
t2
√
Z
u du,
√
sin(2x)
(b) H(x) =
2x
3t2 dt,
Z
x
(c) y(x) =
−x2
t2 dt
Solution: (a) G0 (t) = t2 (4t3 ) − |t|2t = 4t5 − 2t|t|,
(b) H 0 (x) = 3 sin2 (2x)2 cos(2x) − 3(2x)2 2 = 6 sin2 (2x) cos(2x) − 24x2 ,
√
(c) y 0 (x) = x 2√1 x − (−x2 )2 (−2x) = 21 x + 2x5 .
61. The function f (t) is defined in terms of a constant k by


if − 1 ≤ t ≤ 0
k
f (t) = 2k if 0 < t ≤ 1


0
if |t| > 1
Rx
Calculate F (x) = −1 f (t) dt at x = −2, x = − 21 , x = 12 , and for all x > 1.
Solution:
R −2
F (−2) = −1 f (t) dt = 0.
R −1
R −1
F (− 12 ) = −12 f (t) dt = −12 k dt = 21 k.
R0
R1
R 12
f (t) dt = −1 k dt + 02 2k dt = k + k = 2k.
F ( 12 ) = −1
Rx
R0
R1
Rx
If x > 1 then F (x) = −1 f (t) dt = −1 k dt + 0 2k dt + 1 0 dt = k + 2k + 0 = 3k.
62. Sketch the graph of the function
(
x2 + x if 0 ≤ x ≤ 1
f (x) =
2x
if 1 < x ≤ 3.
Rx
Calculate the function F (x) = 0 f (t) dt, with Dom F = [0, 3], and sketch its graph.
Examine the continuity and differentiability properties of both f (x) and F (x) at x = 1.
Solution: For 0 ≤ x ≤ 1, F (x) =
Rx
0
(t2
+ t) dt =
t3
3
+
t2
2
x
=
x3
3
+
x2
2 .
0
9
8
7
6
y
5
4
3
2
f(x)
1
F(x)
0
0
0.5
For 1 < x ≤ 3,
R1
Rx
F (x) = 0 (t2 + t) dt + 1 2t dt =
1
1
3
1.5
x
+
1
2
2
2.5
x
+ t2 =
5
6
3
+ x2 − 1 = x2 − 16 .
1
limx→1− f (x) = limx→1− (x2 + x) = 2 and limx→1+ f (x) = limx→1+ 2x = 2.
Hence f (x) is continuous at x = 1 since limx→1− f (x) = limx→1+ f (x).
limx→1− f 0 (x) = limx→1− (2x + 1) = 3 and limx→1+ f 0 (x) = limx→1+ 2 = 2.
Hence f (x) is not differentiable at x = 1 since limx→1− f 0 (x) 6= limx→1+ f 0 (x).
F (x) is differentiable at x = 1 since F 0 (x) = f (x) is continuous at x = 1.
F (x) is continuous at x = 1 since it is differentiable at x = 1.
63. Evaluate the following definite integrals
Z
−2
(x + 3) dx,
(a)
2
Z
14
6−t
dt,
t3
(b)
−4
1
1
2
Z
8
dq.
1 + 4q 2
(c)
0
Solution:
(a) Put u = x + 3 then
(b)
R2
6−t
1 t3
dt =
R2
1
R −2
(6t−3
3)14 dx
−4
(x +
−
t−2 ) dt
=
=
−
R1
−1
3t−2
u14 du
+
t−1
=
2
u15
15
1
=
−1
= − 34 +
1
2
2
15 .
+ 3 − 1 = 74 .
1
(c) Put 2q = tan u then 2dq =
sec2 u du
and
R
1
2
8
0 1+4q 2
64. Calcuate the following indefinite integrals
Z
Z
x
√
(a)
dx,
(b)
cot x dx,
2 + 3x2
dq =
R π/4
0
Z
(c)
π/4
4 du = 4u
= π.
0
1
√ dx.
1+ x
Solution:
(a) Put u = 2 + 3x2 then du = 6x dx
√
R
R 1 −1/2
1 1/2
1
√ x
dx
=
u
du
=
u
+
c
=
2 + 3x2 + c.
2
6
3
3
2+3x
(b) Put u = sin x then du = cos x dx
R du
R
R
x
cot x dx = cos
sin x dx =
u = log |u| + c = log | sin x| + c.
√
(c) Put u = x then du = 2√1 x dx
R 1
R 2u
R
√
√
2
√ dx =
du
=
(2 − 1+u
) du = 2u − 2 log |1 + u| + c = 2 x − 2 log |1 + x| + c.
1+u
1+ x
65. Calculate the following integrals
Z
Z
√
(a)
x 1 + x dx,
(b)
x2 cos x dx,
Z
(d)
x
e sin(3x) dx,
Z
(e)
Z
(c)
xn log x dx, where n is a positive integer,
e−x sinh x dx.
Solution:
R
R √
4
(1 + x)5/2 + c
(a) x 1 + x dx = 23 x(1 + x)3/2 − 23 (1 + x)3/2 dx = 23 x(1 + x)3/2 − 15
R 2
R
R
(b) x cos x dx = x2 sin x − 2x sin x dx = x2 sin x + 2x cos x − 2 cos x dx
= x2 sin x + 2x cos x − 2 sin x + c.
R n
R
n+1
n+1
n+1
xn+1
(c)
x log x dx = log x xn+1 − x1 xn+1 dx = log x xn+1 − (n+1)
2 + c.
R x
R
(d) e sin(3x) dx = − 31 ex cos(3x) + 31 ex cos(3x) dx
R
= − 13 ex cos(3x) + 91 ex sin(3x) − 91 ex sin(3x) dx and hence
R x
1 x
e sin(3x) dx = 10
e (−3 cos(3x) + sin(3x)) + c.
R
R −x
R
(e)
e sinh x dx = 12 e−x (ex − e−x ) = ( 21 − 12 e−2x ) dx = 12 x + 14 e−2x + c.
66. For integer n ≥ 0 define
xn+2
√
Fn (x) =
dx.
x3 + 1
Find a recurrence relation between Fn (x) and Fn−3 (x) and hence calculate F3 (x) and
F6 (x).
Z
Solution:
√
√
R n+2
R n−1 √
R n−1 x3 +1
3 + 1 dx = 2 xn x3 + 1− 2n
√
Fn (x) = √xx3 +1 dx = 23 xn x3 + 1− 2n
dx
x
x
x
3
3
3
x3 +1
√
2 n
2n
= 3 x x3 + 1 − 3 (Fn (x) + Fn−3 (x)).
√
Hence (3 + 2n)Fn (x) = 2xn x3 + 1 − 2nFn−3 (x).
√
From the first line of this solution we have that F0 (x) = 32 x3 + 1 + c0 .
√
Setting n = 3 in the recurrence relation gives 9F3 (x) = 2x3 x3 + 1 − 6F0 (x) hence
√
F3 (x) = 92 (x3 − 2) x3 + 1 + c3 .
√
Setting n = 6 in the recurrence relation gives 15F6 (x) = 2x6 x3 + 1 − 12F3 (x) hence
√
2
F6 (x) = 45
(3x6 − 4x3 + 8) x3 + 1 + c6 .
67. For integer n ≥ 0 define
Z
π/4
cosn+1 x dx.
In =
0
Find a recurrence relation between In and In−2 and hence evaluate I2 and I4 .
Solution:
R π/4
R π/4
R π/4
π/4
In = 0 cosn+1 x dx = 0 cosn x cos x dx = [cosn x sin x]0 + 0 n cosn−1 x sin2 x dx
R π/4
= √ 1n+1 + n 0 (cosn−1 x (1 − cos2 x)) dx = √ 1n+1 + n(In−2 − In ) hence
2
2
√
In = ( √ 1n+1 + nIn−2 )/(n + 1). From above I0 = 1/ 2 therefore
2
√
√
I2 = ( √1 3 + 2I0 )/3 = 5/(6 2) and I4 = ( √1 5 + 4I2 )/5 = 43/(60 2).
2
2
68. Write the following in partial fraction form
(a)
7x2 − x − 2
,
(x2 − 1)(2x − 1)
(b)
7 − 2x
,
(x + 1)(x − 2)2
(c)
(x −
1)2 (x
8
.
+ 1)(x2 + 1)
Solution:
7x2 − x − 2
A
B
C
=
+
+
2
(x − 1)(2x − 1)
x − 1 x + 1 2x − 1
7x2 − x − 2 7x2 − x − 2 7x2 − x − 2 A=
= 2, B =
= 1, C =
= 1.
(x + 1)(2x − 1) x=1
(x − 1)(2x − 1) x=−1
(x − 1)(x + 1) x=1/2
(a)
A
B
C
7 − 2x
=
+
+
2
(x + 1)(x − 2)
x + 1 x − 2 (x − 2)2
7 − 2x 7 − 2x A=
= 1, C =
= 1,
(x − 2)2 x+1 (b)
x=−1
Evaluating at x = 0 gives
7
4
=1−
B
2
+
x=2
1
4
hence B = −1.
8
A
B
C
Dx + E
=
+
+
+ 2
2
2
(x −
+ 1)(x + 1)
x − 1 (x − 1)
x+1
x +1
8
8
B=
=
2,
C
=
= 1,
(x + 1)(x2 + 1) x=1
(x − 1)2 (x2 + 1) x=−1
(c)
1)2 (x
Evaluating at x = 0 gives E = A + 5.
Evaluating at x = 2 gives 3A + D = −7. Evaluating at x = −2 gives A + 3D = 3.
Hence D = 2, A = −3, E = 2.
−3
2
1
2x + 2
8
=
+
+
+
.
(x − 1)2 (x + 1)(x2 + 1)
x − 1 (x − 1)2 x + 1 x2 + 1
69. Calculate the following integrals
1 + x4
dx,
x2 − x
Z
(a)
Solution:
Z
(b)
x2 + 1
dx,
x(x2 − 1)
Z
(c)
x3
1
dx.
+ x5
1+x
1
2
1 + x4
= x2 + x + 1 + 2
= x2 + x + 1 − +
2
x −x
x −x
x x−1
Z
1
2
1
1
= x2 + x + 1 − +
dx = x3 + x2 + x − log |x| + 2 log |x − 1| + c.
x x−1
3
2
(a)
1 + x4
dx
x2 − x
Z
(b)
x2 + 1
dx =
x(x2 − 1)
Z
Z
−
x2 + 1
1
1
1
=− +
+
2
x(x − 1)
x x−1 x+1
1
1
1
+
+
dx = − log |x| + log |x − 1| + log |x + 1| + c.
x x−1 x+1
(c)
x3
1
A
B
C
Dx + E
= + 2+ 3+ 2
.
5
+x
x
x
x
x +1
1 = C+Bx+(A+C)x2 +(B+E)x3 +(A+D)x4 hence C = 1, B = 0, A = −1, E = 0, D = 1.
Z
Z
1
x
1
1
1
1
dx = − + 3 + 2
dx = − log |x| − 2 + log(x2 + 1) + c.
x3 + x5
x x
x +1
2x
2
70. Evaluate the following integrals
Z
π/4
ex sec2 x
dx,
cosh x
(a)
−π/4
Z
π/3
(b)
−π/3
(1 + x)4 cos x
dx,
1 + 6x2 + x4
Z
1
(c)
−1
1
dx,
1 + esin x
Z
1
(d)
−1
x2
dx.
1 + etan x
Solution: In each case replace the integrand by its even part
Z
π/4
(a)
−π/4
1
ex sec2 x
dx =
cosh x
2
Z
π/4
−π/4
(ex + e−x ) sec2 x
dx =
cosh x
Z
π/4
2
π/4
= 2.
sec x dx = tan x
−π/4
−π/4
π/3
Z π/3
Z π/3
√
(1 + x)4 cos x
(1 + 6x2 + x4 ) cos x
(b)
dx
=
dx
=
cos
x
dx
=
sin
x
=
3.
2
4
1 + 6x2 + x4
−π/3 1 + 6x + x
−π/3
−π/3
−π/3
Z 1
Z
Z
Z
1
1 1
1
1
1 1 2 + esin x + e− sin x
1 1
(c)
dx =
+
dx =
dx =
dx = 1.
sin x
2 −1 1 + esin x 1 + e− sin x
2 −1 2 + esin x + e− sin x
2 −1
−1 1 + e
Z 1
Z
Z
x2
x2
1 1
x2
1 1 2 2 + etan x + e− tan x
dx =
+
dx =
x
dx
(d)
tan x
2 −1 1 + etan x 1 + e− tan x
2 −1
2 + etan x + e− tan x
−1 1 + e
Z 1
1 2
1 3 1
1
=
x dx =
x
= .
6
3
−1 2
−1
Z
π/3
71. Solve the following differential equations for y(x)
(a) 2xyy 0 = 1 + y 2 ,
(b) y 0 + y = 2e−2x ,
(e) (x + y)y 0 = (x − y),
(f) y 0 = (x3 − 2y)/x,
√
(h) y 0 2xy = 1,
(i) 2y 0 + y(yex/2 + 1) = 0.
(d) (xy + xy 3 )dy = log x dx,
(g) 4xyy 0 = x2 + 3y 2 ,
(c) ( xy + 6x)dx + (log x − 2)dy = 0,
Solution:
Z
2y
dy =
1 + y2
Z
(b) I = exp 1 dx = ex ,
Z
1
dx,
x
(a)
(c) M =
∂g
y
= +6x,
∂x
x
y=
y
+ 6x,
x
1
ex
log(1 + y 2 ) = log x + log c,
Z
√
y = ± cx − 1.
ex 2e−2x dx = e−x (−2e−x + c) = −2e−2x + ce−x .
N = log x − 2,
g = y log x+3x2 +φ(y),
∂M
1
∂N
= =
, hence exact.
∂y
x
∂x
∂g
= log x+φ0 = log x−2,
∂y
g = y log x + 3x2 − 2y = c,
y=
φ0 = −2,
φ = −2y.
c − 3x2
.
log x − 2
R
R
(d) (y + y 3 )dy = logx x dx. To integrate the right-hand-side put u = log x with du = x1 dx.
R
1 2
1 4
u du = 12 u2 + c = 12 (log x)2 + c.
2y + 4y =
(e) y 0 = f (x, y) = (x − y)/(x + y), f (xt, yt) = f (x, y) hence homogeneous.
Put y = xv, y 0 = v+xv 0 = (x−xv)/(x+xv) = (1−v)/(1+v), xv 0 = (1−2v−v 2 )/(1+v),
Z
Z
1
1
1
1+v
dv = −
dx,
log(v 2 + 2v − 1) = − log x + log c,
2
v + 2v − 1
x
2
2
v 2 + 2v − 1 = c/x2 , y 2 /x2 + 2y/x − 1 = c/x2 , y 2 + 2xy − x2 = c.
R
(f) y 0 + 2y/x = x2 , I = exp x2 dx = exp(2 log x) = x2 ,
R
y = x12 x4 dx = x12 ( 51 x5 + c) = 15 x3 + xc2 .
(g) y 0 = f (x, y) = (x2 + 3y 2 )/(4xy), f (xt, yt) = f (x, y) hence homogeneous.
Put y = xv, y 0 = v + xv 0 = (x2 + 3y 2 )/(4xy) = (1 + 3v 2 )/(4v), xv 0 = (1 − v 2 )/(4v),
Z
Z
4v
1
dv
=
−
dx, 2 log(v 2 − 1) = − log x + log c2 ,
v2 − 1
x
p
(v 2 − 1)2 = c2 /x, y 2 − x2 = cx3/2 , y = ± x2 + cx3/2 .
p
√
R
R
(h) y 1/2 dy = √12 x−1/2 dx, 23 y 3/2 = 2x + 23 c, y = (3 x/2 + c)2/3 .
(i) Bernoulli with n = 2 so put v = y 1−n = 1/y with y 0 = −v 0 /v 2
R
2v 0
1 1 x/2
− v2 + v v e + 1 = 0, v 0 − 12 v = 21 ex/2 , I = exp − 12 dx = e−x/2 .
R
v = ex/2 12 e−x/2 ex/2 dx = 21 ex/2 (x + c), y = 1/v = 2e−x/2 /(x + c).
72. Solve the following initial value problems for y(x)
(a) (x + yey/x )dx − xey/x dy = 0,
(b) y 0 + 2xy = x,
with y(1) = 0,
with y(0) = 1,
(c) (e2y − y cos(xy))dx + (2xe2y − x cos(xy) + 2y)dy = 0,
(d) xy 0 log x + y = log x,
with y(0) = 1,
with y(e) = 1.
y/x
Solution: (a) y 0 = f (x, y) = x+ye
, f (xt, yt) = f (x, y) hence homogeneous.
xey/x
0
0
Put y = xv, y = v + xv = (x + xvev )/(xev ) = e−v + v, xv 0 = e−v ,
R v
R
e dv = x1 dx, ev = log x + c, v = log(log x + c), y = xv = x log(log x + c),
y(1) = 0 = log c, hence c = 1 giving y = x log(log x + 1).
R
2
2 R
2
2
2
(b) I = exp 2x dx = ex , y = e−x xex dx = e−x ( 12 ex + c),
y(0) = 1 =
1
2
+ c, hence c =
(c) M = e2y −y cos(xy),
hence exact.
1
2
2
to give y = 12 (1 + e−x ).
N = 2xe2y −x cos(xy)+2y,
∂g
= e2y − y cos(xy),
∂x
∂N
∂M
= 2e2y −cos(xy)+xy sin(xy) =
∂y
∂x
g = xe2y − sin(xy) + φ(y),
∂g
= 2xe2y − x cos(xy) + φ0 = 2xe2y − x cos(xy) + 2y,
∂y
φ0 = 2y,
φ = y2.
g = xe2y − sin(xy) + y 2 = c, but y(0) = 1 so c = 1 and xe2y − sin(xy) + y 2 = 1.
Z
1
y
1
0
x
(d) y +
= , I = exp
dx = exp(log(log x)) = log x,
x log x
x
log x
Z
Z
1
1
1
1
1
1
2
y=
u du =
log x dx =
(log x) +c , where u = log x, du = dx,
log x
x
log x
log x 2
x
y(e) = 1 = 1( 12 + c) hence c =
1
2
so y =
1
2
log x +
1
2 log x .
73. Solve the following initial value problems for x(t)
(a) ẋ = 4x,
with x(0) = 2,
(b) 4xẋ = 1,
with x(0) = 1,
(c) tẋ = t2 + 3x,
with x(2) = 12,
Solution:
Z
(a)
(b)
R
dx
=
x
4x dx =
R
Z
dt,
4 dt,
log x = 4t + log c,
x = ce4t ,
x(0) = 2 = c,
2x2 = t + c, but x(0) = 1 hence c = 2, and x =
3
(c) ẋ − x = t,
t
Z
I = exp
x = 2e4t .
q
1 + 12 t.
3
− dt = exp(−3 log t) = t−3 ,
t
R
x = t3 t−2 dt = t3 (−t−1 + c) = −t2 + ct3 ,
x = −t2 + 2t3 .
x(2) = 12 = −4 + 8c, hence c = 2 and
74. Solve the following initial value problems for y(x)
(a) y 00 + 41 y = 0,
y(π) = 1, y 0 (π) = −1.
(b) y 00 − 2y 0 + 5y = 0,
y(π/2) = 0, y 0 (π/2) = 2.
Solution:
(a) (Char) λ2 + 41 = 0, λ = ± 2i , y = A cos(x/2) + B sin(x/2),
y(π) = 1 = B, y 0 = − A2 sin(x/2) + 12 cos(x/2), y 0 (π) = −1 = −A/2, A = 2,
y = 2 cos(x/2) + sin(x/2).
(b) (Char) λ2 − 2λ + 5 = 0, λ = 1 ± 2i, y = ex (A cos(2x) + B sin(2x)),
y(π/2) = 0 = −eπ/2 A, A = 0, y 0 = Bex (sin(2x) + 2 cos(2x)),
y 0 (π/2) = 2 = −2Beπ/2 , B = −e−π/2 , y = −ex−π/2 sin(2x).
75. Solve the following boundary value problems for y(x)
(a) y 00 − 5y 0 + 6y = 0,
y(0) = 1, y(1/2) = e.
(b) y 00 − 2y 0 + 2y = 0,
y(0) = 3, y(π/2) = 0.
Solution:
(a) (Char) λ2 − 5λ + 6 = 0 = (λ − 2)(λ − 3), λ = 2, 3, y = Ae2x + Be3x ,
√
y(0) = 1 = A + B, y(1/2) = e = Ae + Be3/2 , so 1 = A + B e,
giving B = 0, A = 1, y = e2x .
(b) (Char) λ2 − 2λ + 2 = 0, λ = 1 ± i, y = ex (A cos x + B sin x),
y(0) = 3 = A, y(π/2) = 0 = eπ/2 B, so B = 0, giving y = 3ex cos x.
76. Solve the following initial value problems for y(x)
(a) y 00 − 3y 0 + 2y = cos x,
y(0) = 0, y 0 (0) = −1,
(b) y 00 + 2y 0 − 3y = e−x + e2x cos x,
(c) y 00 − 5y 0 + 4y = 3e4x ,
y(0) = 1, y 0 (0) = 1,
y(0) = 0, y 0 (0) = −1,
(d) y 00 + 4y 0 + 4y = −2e−2x + 32e2x ,
y(0) = −1, y 0 (0) = 2.
Solution:
(a) CF: (Char) λ2 − 3λ + 2 = 0 = (λ − 1)(λ − 2), λ = 1, 2, yCF = Aex + Be2x .
PI: y = c cos x+d sin x, −c cos x−d sin x−3(−c sin x+d cos x)+2(c cos x+d sin x) = cos x,
1
3
1
3
c − 3d = 1, d = −3c, c = 10
, d = − 10
, yP I = 10
cos x − 10
sin x.
1
3
1
y = Aex + Be2x + 10
cos x − 10
sin x, y(0) = 0 = A + B + 10
,
3
3
1
sin x − 10
cos x, y 0 (0) = −1 = A + 2B − 10
, A = 12 , B = − 53 ,
y 0 = Aex + 2Be2x − 10
1
3
y = 12 ex − 53 e2x + 10
cos x − 10
sin x.
(b) CF: (Char) λ2 + 2λ − 3 = 0 = (λ − 1)(λ + 3), λ = 1, −3, yCF = Aex + Be−3x .
PI: y = ae−x + e2x (b cos x + c sin x), gives
ae−x + e2x ((3b + 4c) cos x + (3c − 4b) sin x) − 2ae−x + e2x ((4b + 2c) cos x + (4c − 2b) sin x)
−3ae−x + e2x (−3b cos x − 3c sin x) = e−x + e2x cos x,
1
3
1
3
a = − 14 , 6c + 4b = 1, 2c = 3b, b = 13
, c = 26
, yP I = − 41 e−x + e2x ( 13
cos x + 26
sin x),
1 −x
1
3
x
−3x
2x
y = Ae + Be
− 4 e + e ( 13 cos x + 26 sin x),
9
9
, y 0 (0) = 1 = A − 3B + 27
A = 1, B = 52
,
y(0) = 1 = A + B − 52
52 ,
9 −3x
1 −x
1
3
x
2x
y = e + 52 e
− 4 e + e ( 13 cos x + 26 sin x).
(c) CF: (Char) λ2 − 5λ + 4 = 0 = (λ − 1)(λ − 4), λ = 1, 4, yCF = Aex + Be4x .
PI: y = axe4x , ae4x (16x + 8 − 5(4x + 1) + 4x) = 3e4x , a = 1, yP I = xe4x .
y = Aex +Be4x +xe4x , y(0) = 0 = A+B, y 0 (0) = −1 = A+4B+1, A = 32 , B = − 23 ,
y = 32 ex − 32 e4x + xe4x .
(d) CF: (Char) λ2 + 4λ + 4 = 0 = (λ + 2)2 , λ = −2, yCF = e−2x (Ax + B).
PI: y = ax2 e−2x + be2x , ae−2x (4x2 − 8x + 2 − 8x2 + 8x + 4x2 ) + 16be2x = −2e−2x + 32e2x ,
a = −1, b = 2, yP I = −x2 e−2x + 2e2x .
y = e−2x (Ax + B) − x2 e−2x + 2e2x ,
y(0) = −1 = B + 2, y 0 (0) = 2 = A − 2B + 4, A = −8, B = −3.
y = e−2x (−8x − 3) − x2 e−2x + 2e2x .
77. Solve the following pairs of first order differential equations for y(x), z(x) by first finding
a second order differential equation for y(x)
(a)
y 0 = y + 3z,
z 0 = −3y + z,
(b)
y 0 + 6y − 2z = 0,
z 0 + 8y − 4z = 0.
Solution:
(a) z = 13 (y 0 − y), 13 (y 00 − y 0 ) = −3y + 13 (y 0 − y), y 00 − 2y 0 + 10y = 0,
(Char), λ2 − 2λ + 10 = 0, λ = 1 ± 3i, y = ex (A cos(3x) + B sin(3x)).
z = 13 (y 0 − y) = ex (B cos(3x) − A sin(3x)).
(b) z = 21 (y 0 + 6y), 12 (y 00 + 6y 0 ) + 8y − 2(y 0 + 6y) = 0, y 00 + 2y 0 − 8y = 0,
(Char), λ2 + 2λ − 8 = 0 = (λ − 2)(λ + 4), λ = 2, −4, y = Ae2x + Be−4x .
z = 12 (y 0 + 6y) = 4Ae2x + Be−4x .
78. Calculate the second order Taylor polynomial of f (x) = xesin x about x = π/2.
Solution:
f (π/2) = πe
2 ,
0
sin
f (x) = e x (1 + x cos x), f 0 (π/2) = e,
f 00 (x) = esin x (2 cos x − x sin x + x cos2 x),
π
πe
π 2
P2 (x) = πe
2 + e(x − 2 ) − 4 (x − 2 ) .
f 00 (π/2) = − πe
2 ,
79. Calculate the first order Taylor polynomial of f (x) = esin
2
result to show that esin x ≤ 1 + 32 ex2 .
2
x
about x = 0 and use this
Solution:
2
2
f (x) = esin x , f (0) = 1,
f 0 (x) = esin x sin(2x), f 0 (0) = 0,
2
f 00 (x) = esin x (2 cos(2x) + sin2 (2x)), therefore |f 00 (x)| ≤ 3e.
Hence P1 (x) = 1 and f (x) = 1 + R1 (x) where |R1 (x)| ≤ 23 ex2 .
Using the fact that f (x) ≥ 1, the above yields the required result esin
2
x
≤ 1 + 32 ex2 .
80. Use Taylor polynomials to estimate the following to within 0.01
√
(b) sin(0.3),
(c) log(1.2).
(a) e (which is less than 2),
Solution:
1 n
1
x +Rn (x), where Rn (x) = (n+1)!
(a) ex = 1+x+ 12 x2 +· · ·+ n!
ec xn+1 , for some c ∈ (0, x).
√
1 1
1
1
1 √
1
1
1
e = e1/2 = 1 + 21 + 21 212 + · · · + n!
2n + Rn ( 2 ), |Rn ( 2 )| ≤ (n+1)! e 2n+1 ≤ (n+1)! 2n = Qn
1
1
> 0.01, Q3 = 192
< 0.01 so we can use the third order Taylor polynomial
Q2 = 24
√
√
1
1
79
79
1
Note: 48 = 1.6458...,
e ≈ 1 + 2 + 8 + 48 = 48 .
e = 1.6487... .
(b) At x = 0.3 the sine series gives sin(0.3) = 0.3 − 3!1 (0.3)3 + 5!1 (0.3)5 − 7!1 (0.3)7 + . . .
This is a (convergent) alternating series and the magnitude of each term is decreasing.
1
3
As 3! (0.3) = 0.0045 < 0.01 then sin(0.3) ≈ 0.3. Note: sin(0.3) = 0.2955.. .
(c) At x = 1.2 the log series gives
log(1.2) = log(1 + 0.2) = 0.2 − 12 (0.2)2 + 13 (0.2)3 − 14 (0.2)4 + . . .
This is a (convergent) alternating series and the magnitude of each term is decreasing.
As 21 (0.2)2 = 0.02 > 0.01, 13 (0.2)3 = 0.0026.. <0.01, then we need the second
order
Taylor polynomial log(1.2) ≈ 0.2 − 21 (0.2)2 = 0.18.
Note: log(1.2) = 0.1823.. .
81. Let Pn (x) be the nth order Taylor polynomial of the function f (x) = sin x. Find the
least integer n for which
(a) Pn (1) approximates sin(1) to within 0.001
(b) Pn (2) approximates sin(2) to within 0.001
Solution: The Taylor polynomial of sin x of order 2m + 1 is
(−1)m 2m+1
P2m+1 (x) = x − 3!1 x3 + 5!1 x5 + . . . + (2m+1)!
x
.
(a)
(b)
1
5!
29
9!
=
1
120
> 0.001 and
1
7!
=
= 0.0014.. > 0.001 and
1
5040 < 0.001 hence
211
11! = 0.000051.. <
n = 5.
0.001 hence n = 9.
82. Compute the Taylor series of f (x) = x3 + 2x + 1 about x = 2 and comment on the result.
Solution:
f (2) = 13, f 0 (x) = 3x2 + 2, f 0 (2) = 14, f 00 (x) = 6x, f 00 (2) = 12, f 000 (x) = 6.
f (n) (x) = 0 for n ≥ 4.
6
2
3
2
3
f (x) = 13 + 14(x − 2) + 12
2! (x − 2) + 3! (x − 2) = 13 + 14(x − 2) + 6(x − 2) + (x − 2) .
The Taylor series has a finite number of terms and is just a rearrangement of the original
polynomial.
83. Compute the third order Taylor polynomial of f (x) = log(1 + x) about x = 2 and give
the remainder term in Lagrange form.
Solution:
f (2) = log(3), f 0 (x) = 1/(1+x), f 0 (2) = 1/3, f 00 (x) = −1/(1+x)2 ,
f 000 (x) = 2/(1 + x)3 , f 000 (2) = 2/27, f (4) (x) = −6/(1 + x)4
1
1
(x − 2)2 + 81
(x − 2)3 + R3 (x)
log(1 + x) = log(3) + 31 (x − 2) − 18
1
4
where R3 (x) = − 4(1+c)
4 (x − 2) for some c ∈ (2, x).
f 00 (2) = −1/9,
84. Show that esin x = 1 + x + 21 x2 + R2 (x) where |R2 (x)| ≤ 65 e|x|3 .
Solution:
f (x) = esin x , f (0) = 1, f 0 (x) = esin x cos x, f 0 (0) = 1,
f 00 (x) = esin x (− sin x + cos2 x), f 00 (0) = 1,
f 000 (x) = esin x (− cos x − 3 cos x sin x + cos3 x).
esin x = 1 + x + 12 x2 + R2 (x)
|R2 (x)| = | 3!1 esin c (− cos c − 3 cos c sin c + cos3 c)x3 | ≤ 56 e|x|3 .
85. Use Taylor series to calculate limx→0
(2+x)k −2k
,
x
where k is a positive real number.
Solution: From the Taylor series (2 + x)k = 2k + k2k−1 x + o(x) hence
(2 + x)k − 2k
k2k−1 x + o(x)
= lim
= k2k−1 .
x→0
x→0
x
x
lim
86. Calculate the third order Taylor polynomial of f (x) =
√
result to calculate limx→0 1+xf (x)−2x
3 − 1+x3 .
log(1+2x)
1−x
about x = 0 and use this
Solution: log(1 + 2x) = 2x − 12 (2x)2 + 31 (2x)3 + o(x3 ) = 2x − 2x2 + 38 x3 + o(x3 ),
and (1 − x)−1 = 1 + x + x2 + x3 + o(x3 ) hence
f (x) =
log(1+2x)
1−x
= 2x − 2x2 + 83 x3 + 2x2 − 2x3 + 2x3 + o(x3 ) = 2x + 38 x3 + o(x3 ).
√
8 3
8 3
3)
3 ))(1 + x3 + 1 + x3 )
x
+
o(x
(
x
+
o(x
f (x) − 2x
3
√
√
= lim
= lim 3
lim
x→0 1 + x3 − 1 + x3
x→0
x→0 1 + x3 − 1 + x3
(1 + x3 )2 − (1 + x3 )
√
√
( 38 x3 + o(x3 ))(1 + x3 + 1 + x3 )
( 83 + o(1))(1 + x3 + 1 + x3 )
= lim
= lim
= 16/3.
x→0
x→0
x3 + x6
1 + x3
87. Simplify the following expressions involving the positive integer n
+ sin2 ( nπ
) + cos 43π
+ cos 30π
,
(a) sin 55π
2
2
2
2
(b) cos(nπ) + cos(−nπ),
(c)
1
n+2
cos((n + 2)π) +
1
n
cos(nπ).
Solution:
43π
30π
2 nπ
(a) sin 55π
2 + sin ( 2 ) + cos 2 + cos 2
1
3π
1
1
n
n
n
= sin 3π
2 + 2 (1 − (−1) ) + cos 2 + cos π = −1 + 2 (1 − (−1) ) + 0 − 1 = − 2 (3 + (−1) ).
(b) cos(nπ) + cos(−nπ) = 2 cos(nπ) = 2(−1)n .
(c)
1
n+2
cos((n + 2)π) +
1
n
cos(nπ) =
n+n+2
(n+2)n
2(n+1)
n
n(n+2) (−1) .
cos(nπ) =
88. Calculate the Fourier series of
(
1 if − 1 < x < 0
f (x) =
x if 0 ≤ x < 1
and state the value that the series converges to when x = 0.
Solution: The function is defined on (−L, L) with period 2L where L = 1.
Z
1
a0 =
Z
0
f (x) dx =
−1
Z
1 dx +
−1
0
1
3
x dx = .
2
0
1
Z 1
1
1
1
sin(nπx)
+
x sin(nπx) −
sin(nπx) dx
an =
x cos(nπx) dx =
nπ
−1
0
−1 nπ
0 nπ 0
1
1
1
=
cos(nπx) =
((−1)n − 1)
2
2
(nπ)
(nπ)
0
0
1
Z 1
Z 0
Z 1
1
1
1
bn =
sin(nπx) dx+
x sin(nπx) dx = −
cos(nπx)
−
x cos(nπx) +
cos(nπx) dx
nπ
−1
0
−1 nπ
0 nπ 0
1
1
1
1
1
(−1)n +
sin(nπx)
=−
= − (1 − (−1)n ) −
nπ
nπ
(nπ)2
nπ
0
Z
0
Z
cos(nπx) dx+
1
Hence the Fourier series of f (x) is
∞
3 X
+
4
n=1
1
1
n
((−1) − 1) cos(nπx) −
sin(nπx) .
(nπ)2
nπ
f (x) has a jump discontinuity at x = 0 with limx→0− f (x) = 1 and limx→0+ f (x) = 0 hence
at x = 0 the Fourier series converges to 21 (1 + 0) = 12 .
89. Calculate the Fourier series of
(
0
f (x) =
x2
and use this to evaluate the series
P∞
n=1
if − π < x < 0
if 0 ≤ x < π
(−1)n+1
,
n2
P∞
1
n=1 n2 ,
and
P∞
1
n=1 (2n−1)2 .
Solution: The function is defined on (−L, L) with period 2L where L = π.
Z
Z
1 π 2
π2
1 π
f (x) dx =
x dx =
a0 =
.
π −π
π 0
3
π
Z
Z
Z π
1 π 2
1 2
1 π
2
f (x) cos(nx) dx =
x cos(nx) dx =
x sin(nx) dx
an =
x sin(nx) −
π −π
π 0
nπ
0 nπ 0
π
Z
2
1
2
1 π
=−
−
cos(nx) dx = 2 (−1)n
x cos(nx) +
nπ
n
n
n
0
0
π
Z π
Z π
Z π
1
1
2
1
2
2
f (x) sin(nx) dx =
x sin(nx) dx = −
x cos(nx) +
x cos(nx) dx
bn =
π −π
π 0
nπ
0 nπ 0
π
π
Z π
π
π
2
2
2
n
n
= − (−1) + 2 x sin(nx) − 2
sin(nx) dx = − (−1) + 3 cos(nx)
n
n π
n π 0
n
n π
0
0
2
π
(−1)n+1 + 3 ((−1)n − 1)
n
n π
Hence the Fourier series of f (x) is
=
∞
π2 X
+
6
n=1
2
(−1)n cos(nx) +
n2
π
2
n+1
n
(−1)
+ 3 ((−1) − 1) sin(nx) .
n
n π
f (x) is continuous at x = 0 hence at x = 0 the Fourier series converges to f (0) = 0, ie.
∞
π2 X 2
0=
+
(−1)n ,
6
n2
which gives
n=1
∞
X
(−1)n+1
n=1
n2
=
π2
.
12
f (x) has a jump discontinuity at x = π since limx→π− f (x) = π 2 and limx→−π+ f (x) = 0
2
hence at x = π the Fourier series converges to π2 . Therefore
∞
π2
π2 X 2
=
+
2
6
n2
which gives
n=1
∞
X
n=1
∞
X
1
π2
=
.
n2
6
n=1
∞
∞
n=1
n=1
1
1X 1
1 X (−1)n+1
1 π2 1 π2
π2
=
+
=
+
=
(2n − 1)2
2
n2 2
n2
2 6
2 12
8
90. Calculate the Fourier series of


1 if − 2 < x < −1
f (x) = 0 if − 1 ≤ x < 1


1 if 1 ≤ x < 2
Solution: The function is defined on (−L, L) with period 2L where L = 2.
Z 2
Z
1 2
1 dx = 1.
f (x) dx =
a0 =
2 −2
1
Z
Z 2
nπx 2
2
nπ
1 2
nπx
nπx
2
an =
f (x) cos(
cos(
sin(
) dx =
) dx =
) =−
sin( )
2 −2
2
2
nπ
2
nπ
2
1
1
Hence a2k = 0 and a2k+1 =
2
k+1 .
(2k+1)π (−1)
As f (x) is an even function then it has a cosine series (ie. bn = 0)
∞
1 X
2
(2n + 1)πx
.
+
(−1)n+1 cos
2
(2n + 1)π
2
n=0
91. Calculate thePFourier series of f (x) = x3 −π 2 x over (−π, π) and apply Parseval’s theorem
1
to calculate ∞
n=1 n6 .
Solution: As f (x) is an odd function we get a sine series with
π
Z
Z π
1 π 3 2
1
1
3
2
bn =
(x −π x) sin(nx) dx = −
(x −π x) cos(nx)
(3x2 −π 2 ) cos(nx) dx
+
π −π
πn
πn
−π
−π
π
π
Z π
Z π
1
6
6
6
2
2
− 3
=
(3x −π ) sin(nx)
x sin(nx) dx =
x cos(nx)
cos(nx) dx
− 2
πn2
πn3
−π
−π
−π πn
−π πn
12(−1)n
.
n3
∞
X
12(−1)n
sin(nx)
x3 − π 2 x =
n3
=
Hence
n=1
By Parseval’s theorem
Z π
∞
8π 6
1
1 X 144
1 1 7 2 2 5 1 4 3 π
3
2 2
=
(x − π x) dx =
=
x
−
π
x
+
π
x
2π −π
2
n6
2π 7
5
3
105
−π
n=1
hence
∞
X
1
π6
=
.
n6
945
n=1
92. Calculate the half range cosine series of the function f (x) = sin x defined on (0, π).
Solution: For the half range cosine series on (0, L) we work with the even extension of the
function on the interval (−L, L), where L = π.
π
Z
2 π
2
4
a0 =
sin x dx = −
cos x = .
π 0
π
π
0
π
Z π
Z π
2
1
1
a1 =
sin x cos(x) dx =
sin(2x) dx = −
cos(2x) = 0.
π 0
π 0
2π
0
Z
1 π
For n > 1,
sin x cos(nx) dx =
(sin((n + 1)x) − sin((n − 1)x)) dx
π 0
0
π
1
1
1
1 (−1)n + 1 (−1)n + 1
−2((−1)n + 1)
=
−
cos((n+1)x)+
cos((n−1)x) =
−
=
.
π
n+1
n−1
π
n+1
n−1
(n2 − 1)π
0
2
an =
π
Z
π
Hence the half range cosine series of f (x) is
∞
∞
2 X 2((−1)n + 1)
2
4X
1
−
cos(nx)
=
−
cos(2nx).
π
(n2 − 1)π
π π
4n2 − 1
n=2
n=1
93. Calculate the complex Fourier series of
(
−1 if − 2 < x < 0
f (x) =
1
if 0 < x < 2
Solution: The function is defined on (−L, L) with period 2L where L = 2.
1
c0 =
4
For n 6= 0,
Z
1
cn =
4
2
1
f (x) dx = −
4
−2
Z
2
f (x)e
−2
−inπx
2
Z
0
1
1 dx +
4
−2
1
dx = −
4
Z
0
e
Z
2
1 dx = 0.
0
−inπx
2
−2
1
dx +
4
Z
2
e
−inπx
2
dx
0
0
2
−inπx
−inπx
i
i
i
i
2
2
e
e
=−
+
=
(einπ − 1 + e−inπ − 1) =
((−1)n − 1).
2nπ
2nπ
2nπ
nπ
−2
0
2i
and the complex Fourier series of f (x) is
Hence c2n = 0 and c2n+1 = − (2n+1)π
−2i
94. Taylor expand f (x, y) =
quadratic order.
1
1−x−y
∞
X
1
ei(2n+1)πx/2 .
(2n
+
1)π
n=−∞
about (x, y) = (0, 0) up to and including terms of
Solution: f (0, 0) = 1, fx = (1 − x − y)−2 , fx (0, 0) = 1, fy = (1 − x − y)−2 , fy (0, 0) =
1, fxx = 2(1 − x − y)−3 , fxx (0, 0) = 2, fyy = 2(1 − x − y)−3 , fyy (0, 0) = 2, fxy =
2(1 − x − y)−3 , fxy (0, 0) = 2, hence
f (x, y) = 1 + x + y + x2 + 2xy + y 2 + . . .
95. Taylor expand f (x, y) = y 2 /x3 about (x, y) = (1, −1) up to and including terms of
quadratic order.
Solution: f (1, −1) = 1, fx = −3x−4 y 2 , fx (1, −1) = −3, fy = 2x−3 y, fy (1, −1) = −2, fxx =
12x−5 y 2 , fxx (1, −1) = 12, fyy = 2x−3 , fyy (1, −1) = 2, fxy = −6x−4 y, fxy (1, −1) = 6,
hence
f (x, y) = 1 − 3(x − 1) − 2(y + 1) + 6(x − 1)2 + 6(x − 1)(y + 1) + (y + 1)2 + . . .
96. Given f (x, y) = x sin y with x(t) = cos t and y(t) = t2 obtain df
as a function of t using
dt
the chain rule and check the result by expressing f in terms of t and differentiating
directly.
Solution: Using the chain rule
df
∂f dx ∂f dy
dx
dy
=
+
= sin y
+ x cos y
= − sin(t2 ) sin t + cos t cos(t2 )2t.
dt
∂x dt
∂y dt
dt
dt
or directly
f = cos t sin(t2 ),
df
= − sin t sin(t2 ) + 2t cos t cos(t2 ).
dt
97. Given f (x, y) = 4ex log y with x(u, v) = log(u cos v) and y(u, v) = u sin v, obtain ∂f
and
∂u
∂f
as functions of u, v using the chain rule and check the result by expressing f in terms
∂v
of u, v and performing the partial differentiation directly.
Solution: Using the chain rule
∂f
∂f ∂x ∂f ∂y
4ex
4ex
=
+
=
log y +
sin v = 4 cos v log(u sin v) + 4 cos v
∂u
∂x ∂u ∂y ∂u
u
y
∂f
∂f ∂x ∂f ∂y
sin v
4ex
4u cos2 v
=
+
= −4ex log y
+
u cos v = −4u sin v log(u sin v) +
.
∂v
∂x ∂v
∂y ∂v
cos v
y
sin v
or directly
∂f
4u cos2 v
= −4u sin v log(u sin v)+
.
∂v
sin v
∂f
= 4 cos v log(u sin v)+4 cos v,
∂u
f = 4u cos v log(u sin v),
∂z
∂x
98. Given z(x, y) satisfies zx + y log z − z 2 + 4 = 0, calculate
at (x, y, z) = (−3, −1, 1).
Solution: Differentiating the given equation yields
x
99. Calculate
y ∂z
∂z
∂z
+z+
− 2z
= 0,
∂x
z ∂x
∂x
RR
xexy dxdy,
at (x, y, x) = (−3, −1, 1) this gives
∂z
1
= .
∂x
6
where D = [0, a] × [0, b].
D
Solution:
ZZ
xe
Z
xy
xe
dxdy =
Z
a
(e
=
bx
0
RR
xy
Z
a
dy dx =
e
0
0
D
100. Calculate
aZ b
xy
y=b
0
ebx
− 1) dx =
−x
b
a
=
0
dx
y=0
eab − 1
− a.
b
where D = [0, π] × [0, π].
cos(x + y) dxdy,
D
Solution:
ZZ
π
Z
π
Z
cos(x + y) dxdy =
Z
0
π
y=π
sin(x + y)
dx
0
π
y=0
π
(sin(x + π) − sin x) dx =
=
Z
cos(x + y) dy dx =
0
D
101. Calculate
RR
= −4.
− cos(x + π) + cos x
0
0
x3 y dxdy,
where D is the triangle with vertices (0, 0), (1, 0), (1, 1).
D
Solution:
ZZ
3
Z
1Z x
x y dxdy =
0
Z
=
0
102. Calculate
RR √
Z
x y dy dx =
0
D
3
1
1 3 2 y=x
x y
0
2
6 1
x5
x
1
dx =
= .
2
12 0 12
xy dxdy,
D
where D is the finite region between the curves y = x and y = x2 .
y=0
dx
Solution:
√
ZZ
xy dxdy =
RR
1
Z
2
=
3
1
Z
xy dy dx =
0
y=x
2 1 3
x2 y 2
3
dx
y=x2
2
2 x3 2 9 1
2
(x − x ) dx =
= .
− x
3 3
9
27
0
7
2
2
0
√
x2
0
D
103. Calculate
1Z x
Z
ey cosh x dxdy,
where D = {(x, y) : 1 ≤ x ≤ 2, |y| ≤ x}.
D
Solution:
ZZ
1
D
1
=
2
1
Z
2x
(e
−e
1
Z
) dx =
0
0
RR
104. Calculate
−2x
y=x
Z 1 2 y+x
dx
(ey+x + ey−x ) dy dx =
e
+ ey−x
2 1
−x
y=−x
2Z x
Z
1
2
ey cosh x dxdy =
y
√
2
1 − x2 dxdy,
1
sinh(2x) dx =
cosh(2x)
2
1
0
1
= (cosh 4 − cosh 2).
2
where D is the unit disc centred at the origin.
D
Solution:
Z
ZZ
p
2
2
y 1 − x dxdy =
1
√
Z
=
105. Calculate
2
3
Z
RR
1
2
3
(1 − x2 )2 dx =
−1
ex−y dxdy,
Z
p
2
2
y 1 − x dy dx =
√
− 1−x2
−1
D
1−x2
Z
1
−1
y3 p
1 − x2
3
y=√1−x2
√
y=− 1−x2
dx
1
32
2
2
1
= .
x − x3 + x5
3
3
5
45
−1
1
(1 − 2x2 + x4 ) dx =
−1
where D is the triangle with vertices (0, 0), (1, 2), (2, 1).
D
Solution:
ZZ
x−y
e
dxdy =
e
0
D
Z
=
0
1 Z 2y
Z
x=2y
Z
1
ex−y
dy +
x=y/2
2
x−y
Z
dx dy +
y/2
ex−y
Z
dy =
1
dx dy
y/2
(ey − e−y/2 ) dy +
Z
0
x=y/2
x−y
e
1
x=3−y
1
2 Z 3−y
2
(e3−2y − e−y/2 ) dy
1
1 2
1 3−2y
3
1
y
−y/2
−y/2
= e + 2e
+ − e
+ 2e
=
e+
− 3.
2
2
e
0
1
106. Calculate
Z
π/2
Z
0
π/2
x
sin y
dy dx.
y
Solution:
The given iterated integral can be written as a double integral over the region D between the
curves y = x and y = π/2 for 0 ≤ x ≤ π/2, hence
Z π/2 Z π/2
ZZ
Z π/2 Z y
sin y
sin y
sin y
dy dx =
dxdy =
dx dy
y
y
y
0
x
0
0
D
Z π/2 =
0
x sin y
y
x=y
Z
dy =
x=0
π/2
sin y dy =
0
π/2
− cos y
= 1.
0
107. Calculate
Z
1
Z
1
x3
p
dy dx.
x4 + y 2
x2
0
Solution:
The given iterated integral can be written as a double integral over the region D between the
curves y = x2 and y = 1 for 0 ≤ x ≤ 1, hence
ZZ
Z 1 Z √y
Z 1Z 1
x3
x3
x3
p
p
p
dy dx =
dxdy =
dx dy
x4 + y 2
x4 + y 2
x4 + y 2
0
0
0
x2
D
1
=
2
x=√y
1 p
Z
x4
+
y2
0
x=0
1
dy =
2
RR
108. Use polar coordinates to calculate
1
Z
√
( 2 − 1)y dy =
√
0
(x + y) dxdy,
√
2−1 2 1
2−1
y
=
.
4
4
0
where D is the intersection of
D
the half plane x ≤ 0 with the disc of radius 3 centred at the origin.
Solution:
In polar coordinates D is the region r ≤ 3 and π/2 ≤ θ ≤ 3π/2.
ZZ
Z 3 Z 3π/2
r(cos θ + sin θ) dθ rdr
(x + y) dxdy =
0
D
Z
θ=3π/2
3
r2 dr = −
sin θ − cos θ
=
0
π/2
RR
3
2r2 dr = −
0
θ=π/2
109. Use polar coordinates to calculate
Z
e−(x
2 +y 2 )
dxdy,
2 3 3
r
= −18.
3
0
where D is the unit disc centred
D
at the origin.
Solution:
ZZ
Z
−(x2 +y 2 )
e
) dxdy =
1 Z 2π
e
0
D
−r2
Z
dθ rdr = 2π
0
1
−r2
re
−r2
1
dr = −π e
0
0
110. Use the change of variables x = u sin v and y = 12 u cos v to calculate
RR
D
2
1
= π 1− .
e
x2
x2 +4y 2
dxdy,
where D is the region between the two ellipses x2 + 4y 2 = 1 and x2 + 4y = 4.
Solution:
x2 + 4y 2 = u2 , hence D is the region 1 ≤ u ≤ 2 and 0 ≤ v ≤ 2π.
∂x
∂x u cos v
∂u
∂v
sin v
∂(x, y) J=
=
=
1
∂(u, v) ∂y
∂y cos v
− 12 u sin v
2
∂u
∂v
ZZ
x2
dxdy =
x2 + 4y 2
D
Z
=
0
ZZ
u2 sin2 v 1
u dudv =
u2
2
D
2π
1 2
u
4
u=2
u=1
3
sin v dv =
8
2
Z
0
2π
Z
0
2π
Z
1
2
= − 1 u.
2
1
u du sin2 v dv
2
2π
3
1
3π
(1 − cos 2v) dv =
v − sin 2v
=
.
8
2
4
0
111. Use the change of variables x =
1
(u+v)
2
and y =
1
(v−u)
2
to calculate
RR
D
cos
x−y
x+y
dxdy,
where D is the triangle with (x, y) vertices (0, 0), (1, 0), (0, 1).
Solution:
In terms of [u, v] coordinates the vertices of the triangle are [0, 0], [1, 1], [−1, 1]. The edges
of the triangle lie along the lines x = 0, y = 0, y = 1 − x, which become the lines
v = −u, v = u, v = 1, hence D is also a triangle in the uv-plane.
∂(x, y) J=
=
∂(u, v) ZZ
cos
x−y
x+y
ZZ
dxdy =
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
1
1
cos(u/v) dudv =
2
2
D
D
1
=
2
Z
0
u=v
Z
v sin(u/v)
dv =
1
u=−v
1
2
=
1
−
2
0
1
1
2
1
2
Z
0
1
= .
2
1Z v
cos(u/v) du dv
−v
1 2 1 1
v sin(1) dv = sin(1) v
= sin(1)
2
2
0