Written Homework #13 Solutions
MATH 1310 - CSM
Solutions to: PP 412-416; problems: 1, 2abc, 4adg, 8ad, 10, 11bg. PP 695-699; problems:
9, 10abdegikl, 11, 12, 15, 16. PP 708-710; problems: 1abcdefimnpqr, 3, 4, 5.
PP 412-416
1. (a) Obtain a formula for the accumulation function
Z X
5 dx
A(X) =
2
and sketch its graph on the interval 2 ≤ X ≤ 6.
Solution:
Z X
A(X) =
5 dx
2
X
= 5x2
= 5X − 10.
Graph omitted.
(b) is A0 (X) = 5?
Solution: Yes! Since A(X) = 5X − 10, it’s easy to see that A0 (X) = 5.
2. Let f (x) = 2 + x on the interval 0 ≤ x ≤ 5.
(a) Sketch the graph of y = f (x)
Solution: Graph omitted.
(b) Obtain a formula for the accumulation function
Z X
A(X) =
f (x) dx
0
and sketch its graph on the interval 0 ≤ X ≤ 5.
Solution:
Z X
A(X) =
x + 2 dx
0
X
x2
=
+ 2x 2
0
=
1
X2
+ 2X.
2
(c) Verify that A0 (X) = f (X) for every X in 0 ≤ X ≤ 5.
X2
Solution: Since A(X) =
+ 2X, it’s easy to see that A0 (X) = X + 2, and thus
2
A0 (X) = f (x) for every X in 0 ≤ X ≤ 5. (this can be seen by plugging each of
0, · · · , 5 into A0 (X) and f (x) and verifying that they’re equal)
4. Find A0 (X) when
Z
X
(a) A(X) =
X
cos(x) dx. Solution: A0 (X) = cos(x) = cos(X).
0
Z
X
(d) A(X) =
X
cos(t2 ) dt. Solution: A0 (X) = cos(t2 ) = cos(X 2 ).
0
Z
(g) A(X) =
X
X
ln t dt. Solution: A0 (X) = ln t = ln X.
1
8. Express the solution to each of the following initial value problems as an accumulation
function (that is, as an integral with a variable upper limit of integration).
√
(a) y 0 = cos(x2 ), y( π) = 0
Z
Solution: y =
x
√
cos(t2 ) dt.
π
2
(d) y 0 = e−x , y(0) = 0.
Z
Solution: y =
x
2
e−t dt.
0
10. Determine the exact value of each of the following integrals.
Solutions:
Z 7
3 2 5 3 7 26
2
(a)
2 − 3x + 5x dx = 2x − x + x =
2
3
3
3
3
Z 5π
5π
(b)
sin x dx = − cos x 0 = 2
0
Z
5π
(c)
sin(2x) dx =
0
Z
1
(d)
0
Z
6
(e)
1
Z
(f)
0
4
5π
1
sin(2x) 0 = 0
2
1
et dt = et 0 = e − 1
6
dx/x = ln x 1 = ln 6
4
7 2
6 7u − 12u du = u − 2u = −8136
2
0
5
2
2t 1
1
(g)
=
2 dt =
ln 2 0 ln 2
0
Z 1
1 3 1
2
2
s ds = s =
(h)
3
3
−1
−1
Z
1
t
11. Express the values of the following integrals in terms of the parameters they contain.
Solutions:
π
Z π
1
(b)
sin(αx) dx = cos(αx) = 0
α
0
0
Z 1
at 1 a − 1
(g)
at dt =
=
ln a 0
ln a
0
PP 696-699
9. Find an antiderivative of each of the following functions. Don’t hesitate to use the ”trial
and error” method of Example 4 in the text.
Solutions:
Z
3 dx = 3x
Z
5t dt =
Z
−5t dt =
Z
3 − 5t dt
Z
7x4 dx
Z
1
dy
y3
Z
e2z dz
Z
u+
Z
Z
3 2
=
=
=
=
1
du =
u
(1 + w ) dw =
1 + 2w3 + w6 dw =
Z
cos(5v) dv =
Z
x9 + 5x7 − 2x5 dx =
Z
sin t cos t dt =
3
5 2
t
2
5
− t2
2
5
3t − t2
2
7 5
x
5
−1
2y 2
1 2z
e
2
1 2
u + ln u
2
1
1
w + w4 + w7
2
7
1
cos(5v)
5
1 10 5 8 1 6
x + x − x
10
8
3
1
sin2 t
2
10. Find a formula for each of the following indefinite integrals.
Solutions:
Z
3
(a)
3x dx = x2 + c
2
Z
3
(b)
3u du = u2 + c
2
Z
(d)
5t4 dt = t5 + c
Z
(e)
7y +
7
1
dy = y 2 + ln y + c
7
2
Z
5 sin w − 2 cos w dw = −5 cos w − 2 sin w + c
(g)
Z
(i)
cos(4x) dx =
Z
5
dr = 5 arctan r + c
1 + r2
Z
1
1
p
ds = arcsin(2s) + c
2
1 − (2s)2
(k)
(l)
1
sin(4x) + c
4
11. (a) Find an antiderivative F (x) of f (x) = 7 for which F (0) = 12.
Z
Z
Solution:
f (x) dx = 7 dx, which implies that F (x) = 7x + c. Now use the fact
that F (0) = 12. So F (0) = 12 = 0 + c, so c = 12. Hence F (x) = 7x + 12.
(b) Find an antiderivative G(x) of f (x) = 7 for which G(3) = 1
Z
Z
Solution:
f (x) dx = 7 dx, which implies that G(x) = 7x + c. Now use the fact
that G(3) = 1. So G(3) = 1 = 21 + c, so c = −20. Hence G(x) = 7x − 20.
(c) Do F (x) and G(x) differ by a constant? If so, what is the value of that constant?
Solution: Yes! F (x) − G(x) = (7x + 12) − (7x − 20) = 32.
12. (a) Find an antiderivative F (t) of f (t) = t + cos t for which F (0) = 3.
Z
Z
1
Solution:
f (t) dt = t + cos t dt, which implies that F (t) = t2 + sin t + c.
2
Now we can use the fact that F (0) = 3. So F (0) = 3 = 0 + 0 + c, so c = 3 and hence
1
F (t) = t2 + sin t + 3.
2
4
(b) Find an antiderivative
G(t)
Z
Z of f (t) = t + cos t for which G(π/2) = −5
1
Solution:
f (t) dt = t + cos t dt, which implies that G(t) = t2 + sin t + c.
2
Now we can use the fact that G(π/2) = −5:
2
2
π2
π
π
G(π/2) = −5 =
+ sin(π/2) + c. So, c = −
+1+5 =−
+6 .
2 · 22
8
8
2
1
π
Hence G(t) = t2 + sin t −
+6 .
2
8
(c) Do F (t) and G(t) differ by a constant? If so, what is the value of that constant?
2
1 2
1 2
π
Solution: Yes! F (x) − G(x) =
t + sin t + 3 −
t + sin t −
+6
2
2
8
2
π2
π
+6 =
+ 9.
=3−−
8
8
15. (a) Verify that x ln x is an antiderivative of 1 + ln x.
Z
Solution: In order to do this, we need to show that
1 + ln x dx = x ln x. Another
d
x ln x = 1 + ln x, so let’s do that!
dx
d
1
x ln x = (1)(ln x) + (x)
(product Rule)
dx
x
x
= ln x +
x
= 1 + ln x.
way to do this is to show that
d
Since
x ln x = 1 + ln x, if we integrate both sides we see that x ln x =
dx
and thus x ln x is an antiderivative of 1 + ln x.
Z
1 + ln x dx
(b) Find an antiderivative of ln x. [Do you see how you can use part (a) to find this
antiderivative?]
Solution: One possible antiderivative of ln x which we can deduce from part (a) is
d
x ln x − x since
x ln x − x = 1 + ln x − 1 = ln x.
dx
16. Recall that F (y) = ln(y) is an antiderivative of 1/y for y > 0. According to the text, every
antiderivative of 1/y over this domain must be of the form ln(y) + C for an appropriate
value of C.
(a) Verify that G(y) = ln(2y) is also an antiderivative of 1/y.
Solution: We do this in the same way as the previous problem.
d
2
ln(2y) =
dy
2y
1
=
y
Thus G(y) = ln(2y) is an antiderivative of 1/y.
5
(b) Find C so that ln(2y) = ln(y) + C.
Solution: Since ln(2y) can be re-written based on the properties of logarithms as
ln(2) + ln(y), we choose C = ln(2) so that ln(2y) = ln(y) + ln(2).
PP 708-710
1. Evaluate the following integrals using substitution.
Z
(a)
2y(y 2 + 1)50 dy
Solution: Let u = y 2 + 1, then du = 2y dy and the integral becomes:
Z
50
u
u51
du =
+ c. Thus
51
Z
2y(y 2 + 1)50 dy =
(y 2 + 1)51
+ c.
51
Z
(b)
sin(5z) dz
du
= dz and the integral becomes:
Solution: Let u = 5z, then du = 5 dz and
5
Z
Z
1
1
1
sin u du = − cos u + c. Thus sin(5z) dz = − cos(5z) + c.
5
5
5
√
Z
e x
√ dx
(c)
x
√
1
dx
Solution: Let u = x, then du = √ dx, thus 2du = √ and the integral becomes:
2 x
x
Z
Z √x
√
e
√ dx = 2 e x + c.
2 eu du = 2eu + c. Hence
x
Z
(d) (5t + 7)50 dt
1
Solution: Let u = 5t + 7, then du = 5 dt; thus du = dt and the integral becomes:
5
Z
Z
1
u51
(5t + 7)51
50
u du =
+ c. Hence (5t + 7)50 dt =
+ c.
5
5 · 51
5 · 51
Z
p
3
(e)
3u2 u3 + 8 du
3
2
Solution:
Let
the integral becomes:
Z
Z w = u + 8, then dw = 3uZ du and
p
√
3
3
3
3
w dw = w1/3 = u4/3 + c. Hence
3u2 u3 + 8 du = (u3 + 8)4/3 + c.
4
4
Z
1
(f)
dv
2v + 1
du
Solution: Let u = 2v + 1, then
= dv and the integral becomes:
Z
Z 2
1
1
1
1
1
dv = ln u + c. Hence
dv = ln(2v + 1) + c.
2
u
2
2v + 1
2
6
Z
(i)
sec(x/2) tan(x/2) dx
Solution: Let’s re-write the original integral in the following way:
Z
Z
sin x
sec(x/2) tan(x/2) dx =
dx
cos2 x
Now let u = cos x, then du = sin x dx and the integral becomes:
Z
Z
1
1
sin x
−1
du = − + c. Hence
dx =
+ c.
2
2
u
u
cos x
cos x
Z
dr
(m)
r ln r
1
Solution: Let u = ln r, then du = dr and the integral becomes:
r
Z
Z
dr
1
du = ln u + c. Hence
= ln(ln r) + c.
u
r ln r
Z
(n)
ex sin(1 + ex ) dx
Solution:
Let u = 1 + ex , then Zdu = ex dx and the integral becomes:
Z
sin u du = − cos u + c. Hence ex sin(1 + ex ) dx = − cos(1 + ex ) + c.
Z
(p)
w
√
dw
1 − w2
1
Solution: Let u = 1 − w2 , then − du = w dw and the integral becomes:
Z
Z2
p
√
1
w
√
− √ du = − u + c. Hence
dw = − 1 − w2 + c.
2 u
1 − w2
Z
1
(q)
dy
1 + (2y)2
1
Solution: Let u = 2y, then du = dy and the integral becomes:
2
Z
Z
1
1
1
1
1
du = arctan u + c. Hence
dy = arctan(2y) + c.
2
1 + u2
2
1 + (2y)2
2
Z
1
p
dw
(r)
1 − (3w)2
1
Solution: Let u = 3w, then du = dw and the integral becomes:
3
Z
Z
1
1
1
1
1
√
p
du = arcsin u + c. Hence
dw = arcsin(3w) + c.
2
2
3
3
3
1−u
1 − (3w)
Z
3. This question concerns the integral I =
sin x cos x dx.
(a) Find I by using the substitution u = sin x.
Solution: Let u = sin x, then du = cos x dx and the integral becomes:
Z
1
u2
u du =
+ c and hence I = sin2 x + c.
2
2
7
(b) Find I by using the substitution u = cos x.
Solution: Let u = cos x, then − du = sin x dx and the integral becomes:
Z
u2
1
−u du = −
+ c and hence I = − cos2 x + c.
2
2
(c) Compare your answers to (a) and (b). Are they the same? If not, how do they
differ? Since both answers are antiderivatives of sin x cos x, they should differ only
by a constant. Is that true here? If so, what is the constant?
Solution: Our answers are not the same, and so we know they must only differ by
a constant; let’s take their difference:
1
1
1
2
2
sin x + c − − cos x + c =
sin2 x + cos2 x
2
2
2
1
=
2
So our answers differ by the constant
1
.
2
(d) Now calculate the value of the def inite integral
Z
π/2
sin x cos x dx
0
twice, using the two indef inite integrals that you found in (a) and (b). Do the two
values agree, or disagree? Is your result consistent with what you expect?
1
1
Solution: From part (a) we get
and from part (b) we also get , which is not
2
2
what we expected as these two values agree.
4. (a) Find all functions y = F (x) that satisfy the differential equation
dy
= x2 (1 + x3 )13 .
dx
Solution: Using u-substitution, we see that the functions that satisfy the above
differential equation are of the form:
Z
x2 (1 + x3 )13 dx =
1
(1 + x3 )14 + c for some c ∈ R.
42
(b) From among the functions F (x) you found in part (a), select the one that satisfies
F (0) = 4.
1
1
3 14
Solution: y = (1 + x ) + 4 −
42
42
(c) From among the functions F (x), select the one that satisfies F (−1) = 4.
Solution: y =
1
(1 + x3 )14 + 4
42
8
5. Find a function y = G(t) that solves the initial value problem
dy
2
= te−t
dt
y(0) = 3.
Solution: Using separation of variables and u-substitution we see that
Z
Z
2
dy = te−t dt
1 2
y = − e−t + c for some c ∈ R.
2
Now let’s solve for c using the fact that y(0) = 3:
1
y(0) = − e0 + c = 3
2
1
− +c=3
2
7
c=
2
1 2 7
Thus y = − e−t + is a solution to the above initial value problem.
2
2
9