Calculus
Instructor: Mohamed Omar
Assignment 8
Due: Friday, Dec. 7, 2012. 12:00pm
Math 1a Section 1
1. Apostol 6.17 # 24, 27, 30, 33
Solutions:
24) We differentiate each summand with respect to x. Since a is a constant, the
a
a
a
derivative of xa is aa xa −1 . For the second summand, we have ax , and since a
a
a
is constant, the derivative is log(a) · (axa−1 ) · ax = a log(a)xa−1 ax . Finally, in
x
x
the last summand, we have that the derivative of aa is (log(a) · ax ) · log(a)aa .
Thus
a
a
x
f 0 (x) = aa xa −1 + a log(a)xa−1 ax + (log(a))2 ax aa .
27) The book should have defined this function for positive reals x because it’s not
clearly defined for negative reals (for instance, it doesn’t make sense to define
this at x = −1/2) so we will assume x > 0. Then log(f (x)) = xx log(x), so
log(log(f (x))) = log(xx ) + log(log(x)) = x log(x) + log(log(x)). Differentiating
both sides, we get
1
1
1
·
· f 0 (x) = log(x) + 1 +
log(f (x)) f (x)
x log(x)
and hence
0
f (x) =
1
log(x) + 1 +
x log(x)
x
· xx · xx log(x).
30) Again taking logs of both sides, we have
log(f (x)) = log((log(x))x ) − log(xlog(x) ) = x log(log(x)) − (log(x))2 .
Thus
1
2 log(x)
f 0 (x)
= log(log(x)) +
−
,
f (x)
log(x)
x
and hence
(log(x))x
f (x) =
xlog(x)
0
log(log(x)) +
1
2 log(x)
−
log(x)
x
.
33)
log(f (x)) = 2 log(x) +
Thus
1
2
log(3 − x) − log(1 − x) + log(3 + x).
3
3
x2 (3 − x)1/3
f (x) =
(1 − x)(3 + x)2/3
0
1
2
1
1
2
−
−
+
x 9 − 3x 1 − x 9 + 3x
.
2. Apostol 6.17 # 17, 18. Apostol 6.22 # 8, 30, 43, 47.
R
√
6.17, 17) Let u(x) = x. Then u0 (x) = 2√1 x so the integral becomes 2ueu du. We’ve
computed this integral before by parts, and got 2(ueu − eu ), so the integral in
question is
√
√
(2 x − 1)e x .
R
2
6.17, 18) The integral can be expressed as x2 (xe−x ) dx which we can integrate by parts
2
with f = x2 , g 0 = xe−x to get
Z
2
2
2
x2 e−x
e−x
x2 e−x
−x2
+ xe
dx =
−
+C
−2
−2
2
R
R
R
6.22, 8) arccsc(x) dx = arcsin( x1 ) dx so we compute arcsin( x1 ) dx. Let f 0 = 1, g =
arcsin( x1 ). Then by parts
Z
Z
1
−1
1
dx = x arccsc(x) − x · p
· 2 dx
arcsin
2
x
x
1 − (1/x)
Z
1
1
p
= x arccsc(x) +
dx
x 1 − (1/x)2
Z
x
1
= x arccsc(x) +
·√
dx.
2
|x|
x −1
(The sign |x|/x appears when we square x and move
p it inside the square root,
since the square root term is always positive but x 1 − 1/x2 has the same sign
as x.)
6.22, 30) Re-write the integrand by completing the square to get that the integral is
Z
Z
1
1
1
1+x
p
√
dx = √ · r
dx
=
arcsin
+C
2
2
2
2 − (1 + x)2
1+x
1 − √2
6.22, 43) Let u = ex . The u0 (x) = ex so the integral is
Z
1
du = arctan(u) + C = arctan(ex ) + C.
1 + u2
6.22,47) As given in the problem, let x = a+(b−a) sin2 u. Then b−x = (b−a)+(a−x) =
(b − a) − (b − a) sin2 u = (b − a) cos2 u. Thus (x − a)(b − x) = (b − a)2 sin2 u cos2 u.
Since dx = 2(b − a) sin u cos u du this implies our integral is
Z
Z
1
b−a
· 2 · (b − a) · sin u cos u du = 2
du
|(b − a)| sin u cos u
|b − a|
r
b−a
b−a
x−a
=2
u+C =2
arcsin
.
|b − a|
|b − a|
b−a
3. Apostol 6.25 # 1, 5, 16, 38
2
1) If A, B are such that 2x + 3 = A(x + 5) + B(x − 2), then plugging in x = 2 gives
us 7 = 7A and hence A = 1, and plugging in x = −5 gives us −7 = −7B and
hence B = 1. Thus
Z
Z
Z
2x + 3
1
1
dx =
dx +
dx = log |x − 2| + log |x + 5| + C
(x − 2)(x + 5)
x−2
x+5
5) We want to determine A, B, C, D such that
8x3 + 7
A
B
C
D
=
+
+
+
3
2
(x + 1)(2x + 1)
x + 1 2x + 1 (2x + 1)
(2x + 1)3
Simplifying numerators we have
8x3 + 7 = A(2x + 1)3 + B(x + 1)(2x + 1)2 + C(x + 1)(2x + 1) + D(x + 1).
1
Plugging in x = −1
2 we get 6 = 2 D so D = 12. Similarly, plugging in x = −1 we
get −1 = −A and hence A = 1. Thus
8x3 + 7 = (2x + 1)3 + B(x + 1)(2x + 1)2 + C(x + 1)(2x + 1) − 2(x + 1).
Comparing the x3 terms we get 8 + 2B = 8 and hence B = 0. Now comparing
the x2 term we get 12 + 8B + 2C = 0 and hence C = −6. Thus
1
−6
12
8x3 + 7
=
+
+
.
(x + 1)(2x + 1)3
x + 1 (2x + 1)2 (2x + 1)3
Thus the integral is
log |x + 1| +
3
3
−
+ C.
2x + 1 (2x + 1)2
16) We need to determine scalars such that
x−3
A
B
C
= +
+
.
x(x + 1)(x + 2)
x
x+1 x+3
Simplifiying numerators we have
x − 3 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1).
Plugging in x = −1 gives us −4 = −1B so B = 4. Plugging in x = −2 we get
−5 = 2C so C = 25 . Plugging in x = 0 we get −3 = 2A so A = − 32 . Thus
−5
−3
x−3
4
= 2 +
+ 2 .
x(x + 1)(x + 2)
x
x+1 x+2
Hence our integral is
3
5
− log |x| + 4 log |x + 1| − log |x + 3| + C.
2
2
3
38) Re-write the integrand as
√
x2
x
1 2x + 1
1
1
= √
− √
2
2
2 x +x+1 2 x +x+1
+x+1
√
Now integrate each piece. The first summand, we can
substitute
u
=
x2 + x + 1.
√
2
Since du = (2x + 1) dx, we get an antiderivative of x + 2x + 1. For the second
piece, notice that if u = 2x + 1, then x2 + x + 1 = 14 (u2 + 3), and u0 = 2 dx. So
Z
Z
Z
du
du
1
√
q
√
dx =
=
.
x2 + x + 1
u2 + 3
2 1 (u2 + 3)
4
√
3 tan θ and du = 3 sec2 θ,
√
Z
Z √
du
3 sec2 θ dθ
3 sec2 θ dθ
√
p
√
=
=
3 sec θ
u2 + 3
3(tan2 θ + 1)
Z
p
=
sec θdθ = log | tan θ + sec θ| + C = log | tan θ + tan2 θ + 1| + C
Substituting u =
Z
√
r
p
u2
1 + 1| + C = log | √ 2x + 1 + 2 x2 + x + 1 | + C
3
3
p
1
= log |2x + 1 + 2 x2 + x + 1| + log( √ ) + C.
3
u
= log | √ +
3
Absorbing the term log( √13 ) into the contant and adding this to the antiderivative
we found for the first summand, we get that the entire integral is equal to
p
p
1
x2 + x + 1 − log |2x + 1 + 2 x2 + x + 1| + C.
2
4. (a) Prove that a sphere with radius r has volume 43 πr3 .
Solution: Consider the half circle
of radius r plotted in the upper-half plane.
√
This is the function f (x) = r2 − x2 over the interval [−r, r]. We want to
compute the volume of this function when rotated about the x-axis, as this is a
sphere of radius r. The volume is
Z r p
Z r
2
2
2
π·
r −x
dx = π ·
(r2 − x2 ) dx
−r
−r
r
x3
r3
r3
4
2
3
3
=π r x−
=π r −
− π −r +
= πr3 .
3 −r
3
3
3
(b) Determine the volume of the solid obtained by rotating, about the y-axis, the
region in the first quadrant bounded by the curves y = x and y = x2 .
These functions intersect at (0, 0) and (1, 1). Since x > x2 on the interval (0, 1),
using the Washer Method, the volume is
Z 1
Z 1
1 1
π
2
2π
x · x dx − 2π
x · x dx = 2π
−
= .
3 4
6
0
0
4
5. Find the Taylor polynomials (of the indicated degree, and at the indicated point) for
the following functions.
x
(a) f (x) = ee ; degree 3, at 0.
Solution: The derivatives of f are
x
x
x
x
f 0 (x) = ex ·ee , f 00 (x) = ex + (ex )2 ee , f 000 (x) = (ex )2 + (ex )3 ) ee + ex + 2(ex )2 ee .
So
f (0) = e, f 0 (0) = e, f 00 (0) = 2e, f 000 (0) = 5e.
Thus the 3rd degree Taylor polynomial at 0 is
5
e + ex + ex2 + ex3 .
6
1
(b) f (x) = 1+x
2 ; degree 2n + 1, at 0.
(Omitted from the problem set.)
(c) f (x) = cos(x), degree 2n, at
π
2.
Solution: Observe that f 0 (x) = − sin(x), f 00 (x) = − cos(x), f 000 (x) = sin(x), f (4) (x) =
cos(x). Thus, the coefficients cycle between cos(π/2), − sin(π/2), − cos(π/2), sin(π/2),
i.e. 0, −1, 0, 1. So The degree 2n Taylor polynomial of cos(x) centered at π/2 is
π 1 π 3
π 5
π 2n−1
1 (−1)n−1 − x−
+
x−
x−
x−
−
+ ··· +
.
2
3!
2
5!
2
(2n − 1)!
2
6. (a) Find the Taylor polynomial of degree n for the function f (x) = log(1 − x) at
x = 0 and use this to evaluate log( 54 ) with an error of at most 10−4 .
Solution:
First, we calculate the degree n Taylor polynomial of f (x). The first few derivatives are f 0 (x) = −1/(1 − x), f 00 (x) = −1/(1 − x)2 , f 000 (x) = −2/(1 − x)3 , and
using mathematical induction we can show that f (n) (x) = −(n − 1)!/(1 − x)n .
Thus,
f (0) = 0, f 0 (0) = −1, f 00 (0) = −1, f 000 (0) = −2, · · · , f (n) (x) = −(n − 1)!.
Thus the nth degree Taylor polynomial for f (x) = log(1 − x) centered at x = 0
is
x2 x3
xn
−x −
−
− ··· −
.
2
3
n
Our aim is to compute log( 54 ) accurate to 4 decimal places. Observe that log( 54 ) =
f (−1/4). We know that
|f (−1/4) − Pn,0 (−1/4)| ≤ M ·
5
1
· (1/4)n+1
(n + 1)!
where M is the maximum of |f (n+1) (t)| on the interval [0, 14 ]. Since |f (n+1) (t)| =
n+1
. Thus
n!/|1 − t|n+1 , its maximum on the interval is certainly at most n! · 43
n+1
1
4
1
1
|f (−1/4) − Pn,0 (x)| ≤
· (1/4)n+1 =
·
.
n+1 3
n + 1 3n+1
How large must we make n so that
1
1
< 4?
(n + 1) · 3n+1
10
Large enough so that
104 < (n + 1)3n+1 .
Certainly n = 6 is sufficient, since (6 + 1)36+1 = 15309. (Checking n = 5, we see
that n = 6 is the smallest integer satisfying our bound. It turns out that n = 5
is, however, the smallest integer which gives us an approximation of the desired
accuracy—but we can’t see that from the error formula. So the error formula’s
bound gives us an n which works, but is not guaranteed to be the smallest n
which works.) Thus P6,0 (−1/4) is an approximation of log( 54 ) with error less
than 10−4 . The value of the approximation is
−1
1 −1 2 1 −1 3 1 −1 4 1 −1 5
27149
−
−
−
−
−
=
4
2
4
3
4
4
4
5
4
122880
or 0.2231 accurate to 4 decimal places.
(b) Show that for x > 0,
2
3
√
3 1 + x − 1 − x + x ≤ 5x .
3
9
81
Solution:
First, we differentiate f (x) and we see that
f 0 (x) =
1
−2
10
00
000
,
f
(x)
=
,
f
(x)
=
,
3(x + 1)2/3
9(x + 1)5/3
27(x + 1)8/3
Since f (0) = 1, f 0 (0) = 13 and f 00 (0) = −2
9 , this implies that the 2nd degree
Taylor polynomial for f (x) centered at 0 is
1+
x x2
− .
3
9
p
Now observe that the denominator of f 000 (x) has a factor of 3 (x + 1)8 . Now for
8 > 1, since (x + 1)8 is an increasing function and its value at
any x > 0, (x + 1)p
3
x = 0 is 1. Thus (x + 1)8 > 1 if x > 0 and hence f 000 (x) < 10
27 . Thus
|f (x) − P2,0 (x)| ≤
10 |x|3
·
,
27 3!
but this is precisely the statement
2
3
√
3 1 + x − 1 − x + x ≤ 5x .
3
9
81
6