Math 1105: Calculus II (Math/Sci majors)
MWF 11am / 12pm, Campion 235
Written homework 7
§7.5, p. 549(76,86), §7.7, p. 567(53), §7.8, p. 578(64,71,74,76,98)
Problem B. If a, b, and c are real constants, compute the integral
Z
dx
.
2
ax + bx + c
(Hint: There are two cases, depending on whether b2 − 4ac is positive or negative.)
Solution. We complete the square:
b
c
ax + bx + c = a x + x +
a
a
b2
b2
c
b
2
=a x + x+ 2 − 2 +
a
4a
4a
a
!
2
b
b2 − 4ac
=a
x+
−
2a
4a2
2
2
Suppose that b2 − 4ac > 0, so that we have
√
√
2 √ 2
2 !
2 − 4ac
b
b
−
4ac
b
b
b2 − 4ac
b
2
=a x+
−
+
−
x+
.
ax + bx + c = a
x+
2a
2a
2a
2a
2a
2a
This factoring indicates that we can do the above integral using partial fractions: If
ax2
1
=
+ bx + c
x+
A
b
2a
√
−
b2 −4ac
2a
B
+
x+
b
2a
+
√
b2 −4ac
2a
,
then
√
√
b
b2 − 4ac
b
b2 − 4ac
1=A x+
+
+B x+
−
2a
2a
2a
2a
√
b2 − 4ac
b
= (A + B)x + (A + B) + (A − B)
.
2a
2a
This implies that A + B = 0, so that B = −A, and
√
b
b2 − 4ac
1 = (A − A) + (A + A)
2a
√ 2a
2
b − 4ac
,
=A·
a
a
a
and B = − √
. We may now compute:
b2 − 4ac
b2 − 4ac
Z
Z
Z
dx
dx
a
dx
a
√
√
−√
=√
2
2 −4ac
2
b
−4ac
b
ax + bx + c
b2 − 4ac
b2 − 4ac
x + 2a
− b 2a
x + 2a
+ b 2a
√
√
2 − 4ac 2 − 4ac b
a
b
b
b
− log x +
+C
log x +
=√
−
+
2a
2a
2a
2a
b2 − 4ac
√
x + b − b2 −4ac a
2a
2a
√
=√
log +C
2
x + b + b −4ac b2 − 4ac
2a
√2a
2 − 4ac 2ax
+
b
−
a
b
+ C.
√
=√
log b2 − 4ac
2ax + b + b2 − 4ac so that A = √
On the other hand, if b2 − 4ac < 0, then 4ac − b2 > 0, so that we have
!
2
2 √
2 !
2
2
b
b
−
4ac
b
4ac
−
b
ax2 + bx + c = a
x+
−
=a
x+
+
.
2a
4a2
2a
2a
Since we have
Z
1
dx
= 2
2
2
(x − s) + r
r
Z
dx
x−s 2
r
1
= tan−1
r
+1
x−s
r
+C
for any constant r 6= 0, we may compute:
Z
dx
1
=
2
ax + bx + c
a
Z
x+
b
2a
2
dx
√
2
4ac−b2
+
2a
!
b
x
+
1
2a
= ·√
· tan−1 √ 2a2 + C
4ac−b
a
4ac − b2
2a
2ax
+
b
2
−1
√
· tan
=√
+ C.
4ac − b2
4ac − b2
Putting these together, we have
Z




dx
=
ax2 + bx + c 


√ a
b2 −4ac
√ 2
4ac−b2
√
2ax+b−√
b2 −4ac log 2ax+b+ b2 −4ac + C,
−1
tan
√2ax+b
4ac−b2
Z
Problem C. Same as above, but compute
ax2
+ C,
if b2 − 4ac > 0, and
if b2 − 4ac < 0.
x dx
.
+ bx + c
Solution. Note that (ax2 + bx + c)0 = 2ax + b, so that
Z
2ax + b
dx =
ax2 + bx + c
Z
du
= log |u| + C = log |ax2 + bx + c| + C.
u
With this in mind, we can rewrite the integrand:
x
1
2ax
=
· 2
2
ax + bx + c
2a ax + bx + c
1 2ax + b − b
·
=
2a ax2 + bx + c
1
2ax + b
b
1
=
· 2
−
· 2
.
2a ax + bx + c 2a ax + bx + c
Now we may use Problem B to continue:
Z
Z
Z
x
1
2ax + b
b
dx
dx =
dx −
2
2
2
ax + bx + c
2a
ax + bx + c
2a
ax + bx + c
Z
b
dx
1
log |ax2 + bx + c| −
=
2a
2a
ax2 + bx + c

√
2ax+b−√
b
b2 −4ac 
√

 2 b2 −4ac log 2ax+b+ b2 −4ac + C,
1
2
log |ax + bx + c| −
=

2a

 √ b 2 tan−1 √2ax+b 2 + C,
a 4ac−b
4ac−b
Z
p. 549, Problem 76. Compute
dx
√ using x = u3 .
x− 3x
Solution. Using the substitution x = u3 and dx = 3u2 du, we have
Z
Z
Z
3u du
dx
3u2 du
√
=
.
=
3
3
u −u
u2 − 1
x− x
We perform a partial fraction decomposition of the integrand:
3u
3u
A
B
=
=
+
,
2
u −1
(u − 1)(u + 1)
u−1 u+1
and clearing denominators we have
3u = A(u + 1) + B(u − 1) = (A + B)u + (A − B).
This implies that A − B = 0 and A + B = 3, or A = B and 2A = 3. Thus
3u
3/2
3/2
=
+
,
u2 − 1
u−1 u+1
and we may compute:
Z
Z
Z
dx
3
du
3
du
√
=
+
3
2
u−1 2
u+1
x− x
3
3
= log |u − 1| + log |u + 1| + C
2
2
3
= log |u2 − 1| + C
2
3
= log |x2/3 − 1| + C.
2
Z
p. 549, Problem 86. Compute
x3 + 1
dx.
x(x2 + x + 1)2
if b2 − 4ac > 0, and
if b2 − 4ac < 0.
Solution. The partial fraction decomposition becomes:
x3 + 1
Bx + C
Dx + E
A
+
= + 2
.
2
2
x(x + x + 1)
x x + x + 1 (x + 2 + x + 1)2
Clearing denominators, we find
x3 + 1 = A(x2 + x + 1)2 + (Bx + C)x(x2 + x + 1) + (Dx + E)x.
Setting x = 0, we find 1 = A. Expanding the above, we have
x3 + 1 = (x2 + x + 1)(x2 + x + 1) + (Bx + C)(x3 + x2 + x) + Dx2 + Ex
= x4 + 2x3 + 3x2 + 2x + 1 + Bx4 + (B + C)x3 + (B + C)x2 + Cx + Dx2 + Ex
= (1 + B)x4 + (B + C + 2)x3 + (B + C + D + 3)x2 + (C + E + 2)x + 1.
This implies that 1 + B = 0, B + C + 2 = 1, B + C + D + 3 = 0, and C + E + 2 = 0. Thus B = −1,
C = 0, D = −2, and E = −2. Thus,
1
−x
−2x − 2
x3 + 1
= + 2
+ 2
.
2
2
x(x + x + 1)
x x + x + 1 (x + x + 1)2
Thus the integral in question is given by
Z
Z
Z
Z
x3 + 1
dx
−x
2x + 2
dx =
+
dx −
dx
2
2
2
2
x(x + x + 1)
x
x +x+1
(x + x + 1)2
The first integral is easy, the second can be computed using Problem C, but the last is a bit harder. Let’s
compute the middle integral before dealing with the harder integral: Since b2 − 4ac = 12 − 4(1)(1) =
−3 < 0, we may read from Problem C that
Z
x
1
1
2x + 1
2
−1
√
dx = log |x + x + 1| − √ tan
+ C.
x2 + x + 1
2
3
3
Towards the last integral, the good news is that the substitution u = x2 + x + 1 seems useful, but the
problem is that du = 2x + 1 dx. In order to make this useful, we break the integral up:
Z
Z
Z
2x + 2
2x + 1
1
dx =
dx +
dx.
2
2
2
2
2
(x + x + 1)
(x + x + 1)
(x + x + 1)2
The first integral may be computed with substitution:
Z
Z
2x + 1
du
1
1
dx =
=− +C =− 2
+ C.
2
2
2
(x + x + 1)
u
u
x +x+1
In order to compute the last integral, following the hint, we’d like to change the integrand into something
of the form 1/(w2 + r2 )2 . In this case, we have:
1
1
1
=
=
2 .
2
(x2 + x + 1)2
1 2
3
x2 + x + 41 + 34
x+ 2 + 4
Changing variables to w = x + 12 , so that dw = dx, we have
Z
Z
Z
dx
dw
=
2 =
(x2 + x + 1)2
w2 + 3
4
dw
2
3 2
2w
4
√
3
2
+1
√
√
Using the substitution 2w/ 3 = tan θ so that dw = 23 sec2 θdθ, we have:
√
Z
Z
3
sec2 θ
dw
2
=
dθ
2
2
3 2
2
(tan
θ
+
1)
w2 + 43
√ 4 2Z
√ Z
√ Z
3 4
sec2 θ
8 3
dθ
8 3
=
· 2
dθ =
=
cos2 θ dθ
4
2
2 3
sec θ
9
sec θ
9
√ √ Z
4 3
sin 2θ
8 3
1 + cos 2θ
dθ =
θ+
+ C.
=
9
2
9
2
In order to substitute back in for w, we need to compute that
2 sin θ cos θ
sin θ
1
tan θ
= sin θ cos θ =
· cos2 θ = tan θ ·
=
.
2
2
cos θ
sec θ
1 + tan2 θ
√
Since 2w/ 3 = tan θ, we have
√
√
Z
2w
√
dw
2w
4
3
4 3
3
−1
√
+
tan
·
=
2 + C
3 2
2
9
9
3
2w
w +4
1+ √
3
√
4 3
2w
8w
=
tan−1 √
+ C.
+
9
3(3 + 4w2 )
3
Returning to x, we have
√
Z
1
8
x
+
2
1
dx
4 3
2
=
tan−1 √ x +
+ 2 + C
(x2 + x + 1)2
9
2
3
3 3 + 4 x + 12
√
4 3
2x + 1
8x + 4
−1
√
=
tan
+
+C
2
9
3(4x + 4x + 4)
3
√
4 3
2x + 1
2x + 1
−1
√
tan
+ C.
=
+
9
3(x2 + x + 1)
3
Putting all of the pieces together, we have:
Z
Z
Z
Z
x3 + 1
dx
x
2x + 2
dx =
−
dx −
dx
2
2
2
2
x(x + x + 1)
x
x +x+1
(x + x + 1)2
Z
Z
Z
Z
dx
x
2x + 1
dx
=
−
dx −
dx −
2
2
2
2
x
x +x+1
(x + x + 1)
(x + x + 1)2
1
2x + 1
1
1
√
= log |x| − log |x2 + x + 1| + √ tan−1
+ 2
2
x +x+1
3
3
√
4 3
2x + 1
2x + 1
−
tan−1
−
+C
2
9
9
3(x + x + 1)
√
1
x2
2 − 2x
3
2x + 1
−1
= log 2
+
−
tan
+ C.
2
x + x + 1 3(x2 + x + 1)
9
9
p. 567, Problem 53. The length of an ellipse with axes of length 2a and 2b is
Z 2π p
a2 cos2 t + b2 sin2 t dt.
0
Use numerical integration and experiment with different values of n to approximate the length of the
ellipse with a = 4 and b = 8.
Solution. We simplify the integral in question slightly:
Z
2π
p
2π
Z
2
42 cos2 t + 82 sin t dt =
0
√
r
42
cos2 t +
0
Z
2π
=4
82
sin2 t dt
2
4
p
1 − sin2 t + 4 sin2 t dt
0
Z
=4
2π
p
1 + 3 sin2 t dt.
0
We perform Simpson’s method for approximation, using f (x) =
make approximations, and list the data:
x
0.0
1.047
2.094
3.142
4.189
5.236
6.283
√
1 + 3 sin2 x, n = 6 and n = 8. We
f(x)
1.0
1.803
1.803
1.0
1.803
1.803
1.0
Simpson’s approximation with n = 6 may now be applied:
S(6) =
1.047
(1.0 + 4(1.803) + 2(1.803) + 4(1.0) + 2(1.803) + 4(1.803) + 1.0) = 9.645.
3
With n = 8 we have:
x
0.0
.785
1.571
2.356
3.142
3.927
4.712
5.498
6.283
f(x)
1.0
1.581
2.0
1.581
1.
1.581
2.0
1.581
1.0
Simpson’s approximation with n = 8 may now be applied:
S(8) =
.785
(1.0 + 4(1.581) + 2(2.0) + 4(1.581) + 2(1.0) + 4(1.581) + 2(2.0) + 4(1.581) + 1.0) = 9.759.
3
In fact, here is a table of more computations of Simpson’s approximation:
n
8
10
12
14
16
18
20
S(n)
9.76465
9.68641
9.70197
9.68832
9.69132
9.68844
9.68912
Error
7.6 × 10−2
2.0 × 10−3
1.3 × 10−2
1.3 × 10−4
2.9 × 10−3
9.9 × 10−6
6.7 × 10−4
Z
∞
p. 578, Problem 64. For what values of p does the integral
2
dx
exist, and what is its value
x logp x
(in terms of p)?
Solution. We perform the substitution u = log x, so that du = dx/x. We have:
Z ∞
Z a
Z log a
dx
dx
du
= lim
= lim
.
p
p
x log x a→∞ 2 x log x a→∞ log 2 up
2
If p = 1, then this becomes
Z ∞
dx
a
lim log(log a) − log(log 2).
= lim log u|log
p
log 2 = a→∞
a→∞
x
log
x
2
Let b = log a, so that, as a goes to ∞, b goes to ∞ as well. Then
lim log(log a) = lim log b = ∞,
a→∞
b→∞
so that the integral doesn’t exist. If p 6= 1, then the integral above becomes
1−p log a
Z ∞
(log a)1−p (log 2)1−p
dx
u
=
lim
=
lim
−
x logp x a→∞ 1 − p log 2 a→∞ 1 − p
1−p
2
b1−p
(log 2)1−p
−
,
b→∞ 1 − p
1−p
where we’ve used b = log a, which goes to ∞ as a goes to ∞. If 1 − p < 0, this last limit exists, and is
equal to
(log 2)1−p
1
−
=
,
1−p
(p − 1)(log 2)p−1
whereas if 1 − p > 0, this last limit doesn’t exist, so neither does the integral.
= lim
p. 578, Problem 71. Let R be the region bounded by the graphs of y = e−ax and y = e−bx , for
x ≥ 0, where a > b > 0. Find the area of R.
Solution. Since a > b, we have eax > ebx for x > 0 (technically, this is because ax > bx for x > 0
and ex is increasing—which holds because (ex )0 = ex > 0 for all x). This implies that e−ax < e−bx for
x > 0, so that the area between these curves is given by
−bx
c
Z ∞
Z c
e
e−ax −bx
−ax
−bx
−ax
−
e
−e
dx = lim
e
−e
dx = lim
c→∞ 0
c→∞
−b
−a 0
0
−bc
e
e−ac
1 1
1 1
= lim −
+
− − +
= − .
c→∞
b
a
b a
b a
p. 578, Problem 74. ConsiderZthe family of functions f (x) = 1/xp , where p is a real number. For
1
f (x) dx exist? What is its value?
what values of p does the integral
0
Solution. We have:
Z
1
x
0
−p
1
Z
dx = lim+
a→0
x−p dx.
a
If −p = −1 (i.e. p = 1), then this integral becomes
1
lim+ (log |x|)|1a = lim+ (− log |a|) = lim+ log .
a→0
a→0
a→0
a
Using a new variable b = 1/a, (where b goes to ∞ as a goes to 0), this limit becomes
lim log b = ∞,
b→∞
so the integral doesn’t exist. If −p 6= −1, then the integral is given by:
1−p 1
Z 1
1−p
1
a
x
−p
= lim
lim
−
x dx = lim+
a→0+ a
a→0
1 − p a a→0+ 1 − p 1 − p
If 1 − p > 0 (i.e. p < 1), then the above limit exists and equals 1/(1 − p), whereas if 1 − p ≤ 0 (i.e.p ≥ 1)
the above limit is infinite (i.e. doesn’t exist).
p. 578, Problem 76. Let R be the region bounded by the graph of f (x) = x−p and the x-axis, for
x ≥ 1.
(a) Let S be the solid generated when R is revolved about the x-axis. For what values of p is the volume
of S finite?
(b) Let S be the solid generated when R is revolved about the y-axis. For what values of p is the volume
of S finite?
Solution (a). The volume of S is given by the integral
Z ∞
Z ∞
−p 2
π x
x−2p dx.
dx = π
1
1
If −2p = −1 (i.e. p = 1/2), then this integral becomes
Z ∞
dx
π
= π lim log a,
a→∞
x
1
and the volume is infinite. If p 6= 1/2, we have:
Z ∞
Z
−2p
πx
dx = π lim
1
a
x1−2p x
dx = π lim
a→∞ 1
a→∞ 1 − 2p 1
1−2p
a
1
= π lim
−
a→∞ 1 − 2p
1 − 2p
a
−2p
If 1 − 2p < 0 (so 2p − 1 > 0), then this limit converges:
a−(2p−1)
1
1
1
π
+
= π lim
+
=
.
2p−1
a→∞ 1 − 2p
a→∞ (1 − 2p)a
2p − 1
2p − 1
2p − 1
π lim
On the other hand, if 1 − 2p > 0, then this limit goes to infinity, and the volume is infinite. That is, if
p > 1/2 the volume is finite, but if p ≤ 1/2 the volume is infinite.
Solution (b). Using the cylindrical shell method, the volume of this solid is given by the integral
Z 1
Z 1
Z 1
−p
1−p
2πx(x ) dx = 2π
x
dx = 2π lim+
x1−p dx.
0
a→0
0
a
From problem 74 on p. 578 above, if p − 1 < 1 (i.e. p < 2) then the above limit exists and equals
2π/(2 − p), whereas if p − 1 ≥ 1 (i.e. p ≥ 2) the above limit is infinite.
p. 578, Problem 98. The gamma function is defined by
Z ∞
xp−1 e−x dx,
Γ(p) =
0
for p not equal to zero or a negative integer.
(a) Use the reduction formula
Z ∞
Z
p −x
x e dx = p
0
∞
xp−1 e−x dx,
0
for p a positive integer, to show that Γ(p + 1) Z
= p!.
∞
2
√
π
du =
to show that Γ
2
−u2
e
(b) Use the substitution x = u and the fact that
0
1
2
=
√
π.
Solution (a). We have:
∞
Z
p −x
xe
Γ(p + 1) =
Z
∞
dx = p
xp−1 e−x dx = p · Γ(p).
0
0
Applying this rule repeatedly to the positive integer p + 1, we find:
Γ(p + 1) = p · Γ(p) = p(p − 1) · Γ(p − 1) = p(p − 1)(p − 2) · Γ(p − 2)
= p(p − 1)(p − 2)(p − 3) · Γ(p − 3) = . . . = p(p − 1) . . . (2)(1) · Γ(1)
= p! · Γ(1).
Note that we have:
Z
∞
Γ(1) =
0
1−1 −x
x
Z
e
∞
dx =
e−x dx
0
a
= lim
a→∞
Z
a
e−x dx = lim − e−x 0
a→∞
0
1
= 1.
ea
Plugging this into the above equation, we have Γ(p + 1) = p!, as desired.
= lim 1 −
a→∞
Solution (b). We deal first with the relevant indefinite integral, using the substitution x = u2 , so
that dx = 2udu:
Z
Z
Z
Z
1 −u2
2
−1/2 −x
2 −1/2 −u2
x
e dx =
u
e
2u du = 2 u · e
du = 2 e−u du.
u
Thus we have:
Z ∞
Z a
Z 1
1
−1/2 −x
−1/2 −x
Γ
=
x
e dx = lim
x
e dx + lim+
x−1/2 e−x dx
a→∞
b→0
2
0
1
b
Z √a
Z 1
2
2
= lim 2
e−u du + lim+ 2 √ e−u du
a→∞
b→0
1
b
Z c
Z 1
−u2
−u2
= 2 lim
e
du + lim+
e
du .
c→∞
1
d→0
d
In the last line above we’ve replaced a with c2 and b with d2 . (Note that as a and b go to ∞, both c
and d go to ∞ as well). The last expression is just:
√ Z ∞
√
1
π
−u2
Γ
=2
e
du = 2
= π,
2
2
0
by the fact that is given.