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.