Integration
1. Find
∞
Z
1
2. Find
Z
√
e−
√
x
x
dx.
(log x)3 dx,
by integrating by parts 3 times.
3. Using L’Hospital’s Rule or otherwise, show that x log x → 0 as x → 0+ , and hence calculate the
improper integral
Z 1
log x dx.
0
4
2
2
∗4. Find A and B if x + 1 = (x + Ax + 1)(x + Bx + 1), and hence find
Z
1
dx.
4
x +1
∗5. Show that
Z
0
π
π2
x sin x
dx
=
.
1 + cos2 x
4
∗∗6. (From Spivak) Calculate
Z
(x2
x2 − 1
√
dx,
+ 1) x4 + 1
by using a substitution.
Approximate integration, surface area and arc length
7. Use Simpson’s rule with n = 2 to approximate the area between the curve y = 4 − x2 and the x-axis.
Compare this with the exact area. Approximate the same area using the trapezoidal rule with n = 2.
∗8. Find the surface area of the figure obtained by rotating y = 1 − x2 around the x-axis between x = −1
and x = 1.
√
√
9. Find the length of the curve y = x − x2 + arcsin x, where 0 ≤ x ≤ 1.
10. Find the length of y = ln(sec x) where 0 ≤ x ≤ π/4.
∗11. Sketch the ellipse
y2
= 1,
2
and use the arc length formula to find a formula for its perimeter (do not evaluate the integral).
x2 +
Find the surface area of the solid formed when the ellipse is rotated about the x-axis (you can evaluate
the final integral using a substitution).
1
Sequences and series
12. Find
√
√
lim ( n + 1 − n).
n→∞
13. Find
lim (ln(n + 1) − ln n).
n→∞
14. Find
lim e1/n .
n→∞
15. Find
n2
4
3 +arctan n
n
lim
.
arctan e
n→∞ π
√
∗16. Let a0 = 1 and let for n ≥ 0 let an+1 = 1 + an .
(a) Show that an is a bounded, increasing sequence, and state why it therefore has a limit L.
(b) Find L.
17. Find
∞
X
1
,
n(n
− 1)
n=2
by using partial fractions.
18. Show that
∞
X
1
ln 1 +
,
n
n=1
diverges.
19. Use the integral test to show that
P∞
1
n=1 n2
converges, and use a comparison to show that
∞
X
sin2 n
,
1 + n2
n=1
converges.
20. Use the integral test to show that the harmonic series
1+
1 1
+ + ...,
2 3
diverges.
21. Show that the harmonic series diverges without using the integral test, by showing that the bracketed
terms in
1 1
1
1 1 1 1
1 1
1
1 + + + ... = 1 +
+
+ + +
+
+ + ... +
+ ...,
2 3
2
3 4 5 6
7 8
14
are all greater than 12 .
2
22. Let
an =
1
1+
n
n
.
Find limn→∞ ln an , and hence find limn→∞ an .
∗23. Show
P∞ that if an and bn are non-negative for all n, and if
n=1 an bn converges.
P∞
n=1
a2n and
P∞
2
n=1 bn
both converge, then
24. (From Rudin) Find
lim
p
n→∞
n2 + n − n .
25. Show that limn→∞ n1/n = 1.
Alternating series, absolute convergence and power series
P∞
26. State what it means for n=1 an to converge absolutely, and to converge conditionally. State the ratio
test, the root test, and the alternating series test.
P∞
n)2
√
converges, and specify if the convergence is condi27. State whether or not the series n=1 (−1)n (log
n
tional or absolute. Justify your answer.
28. Use the ratio test to determine the convergence or divergence of
∞
X
n!
p
n=1
(2n)!
,
and
∞
X
(2n)!
p
.
(n2 )!
n=1
29. Find a power series representation of
Z
f (x) =
1
dx,
1 + x4
30. Find a power series for log(1 + x). For a > 0, find a power series for log(a + x), and state the radius
of convergence in terms of a.
P∞
P∞
2
1
∗31. By differentiating 1−x
as many times as necessary, find the value of n=1 3nn and n=1 3nn .
32. Using
x3
x5
x7
+
−
+ ...,
3!
5!
7!
find a power series representation for cos x around x = 0, and a power series for sin x around x = π.
sin x = x −
∗33. Differentiate
f (x) = 1 + x + x2 /2! + x3 /3! + . . . ,
and hence show that f (x) = ex . Use this to find a power series for (ex + e−x )/2.
R 1
34. Use the formula arctan x = 1+x
2 dx to find a power series for arctan x around x = 0. Find the radius
of convergence, and the interval of convergence for the resulting power series. Using arctan 1 = π/4,
state how many terms are required to estimate π to the nearest whole number, and to 1 decimal place.
3
Misc (all topics)
∗35. Show that
Z
1
log x
−π 2
dx =
.
1+x
12
0
You may assume that
∞
X
π2
1
=
.
2
n
6
n=1
Partial answers
1. −2/e
2. x(log x)3 − 3x(log x)2 + 6x log x − 6x + C
3. -1
4. A =
√
√
2, B = − 2,
√ √
√
√
√
2
log(x2 + 2 + 1) − log(x2 − 2x + 1) + 2 arctan(x 2 + 1) − 2 arctan(x 2 − 1)
8
5. Change variables using u = x − π/2 and use the fact that the integral of an odd function over a
symmetric interval is zero:
Z
π/2
−π/2
x sin x
dx =
1 + cos2 x
Z
π
0
Z
(u + π/2) sin(u + π/2)
du
1 + cos2 (u + π/2)
π/2
=
−π/2
Z
π/2
=
−π/2
Z
(u + π/2) cos u
du since sin(π/2 − t) = cos t,
1 + sin2 u
Z π/2
u cos u
(π/2) cos u
du
+
du
2
1 + sin2 u
−π/2 1 + sin u
π/2
=0+
−π/2
Z 1
(π/2) cos u
du
1 + sin2 u
1
dw
1
+
w2
−1
1
= (π/2) · arctan w
= (π/2) ·
cos(π/2 − t) = sin t
(integral of odd function over symmetric region is zero)
(by substituting w = sin u)
−1
= π 2 /4.
4
6.
Z
x2 1 − x12
dx
q
x2 x + x1
x2 + x12
Z
1 − x12
dx
=
q
2
x + x1
x + x1 − 2
Z
1
√
=
du (substitute u = x + 1/x )
u u2 − 2
Z
√
1
√ dw (substitute u = 2 sec w )
=
2
w
= √ +C
2
√ !
1
2
= √ · arccos
+C
u
2
√ !
1
x 2
= √ · arccos
+C
x2 + 1
2
x2 − 1
√
dx =
(x2 + 1) x4 + 1
Z
7. Simpson’s rule gives the exact area 32/3. The trapezoidal rule gives 8.
√
√ π
14 5 + 17 ln(2 + 5)
8. 16
9. 2
10. ln(1 +
√
2)
11. The perimeter is 2
12.
R 1 q 1+x2
1−x2
−1
√ √
√ dx, the surface area is 2π 2 2 + ln(1 + 2) .
√
√
√
√
lim ( n + 1 − n) = lim ( n + 1 − n) ·
n→∞
n→∞
13.
lim (ln(n + 1) − ln n) = lim ln
n→∞
n→∞
√
√ n+1+ n
1
√
= lim √
√
√ = 0.
n→∞
n+1+ n
n+1+ n
n+1
n
= ln
n+1
n→∞
n
lim
= ln(1) = 0.
14. 1
15. 1
16. L =
√
1+ 5
2 .
17. 1
18.
N
X
1
ln 1 +
n
n=1
=
N
X
n=1
=
N
X
ln
n+1
n
ln (n + 1) − ln(n)
n=1
= (ln(2) − ln(1)) + (ln(3) − ln(2)) + . . . + (ln(N ) − ln(N − 1)) + (ln(N + 1) − ln(N ))
= ln(N + 1)
→ ∞ as N → ∞.
5
19. For every n ≥ 1,
1
1
sin2 n
≤
≤ 2.
1 + n2
1 + n2
n
R∞
P∞
The function x12 is positive and decreasing for x ≥ 1, and 1 x12 dx converges, hence n=1
P∞ sin2 n
by the integral test, and so n=1 1+n2 converges by the comparison test.
20.
21. Each of the bracketed terms is of the form
1
1
1
1
1
1
+
+ ... +
≥
+
+ ... +
k k+1
2k
2k 2k
2k
k+1
=
2k
k
≥
2k
1
= .
2
22. By L’Hospital’s Rule,
lim
x→0
1
log(1 + x)
= lim
= 1,
x→0
x
1+x
hence
n 1
lim ln an = lim log
1+
n→∞
n→∞
n
1
= lim n log 1 +
n→∞
n
log 1 + n1
= lim
n→∞
1/n
= 1.
Hence
lim an = lim elog an = e1 = e.
n→∞
n→∞
23. Use
ab ≤
a2 + b2
,
2
and the comparison test.
24.
1
2.
25. Use n = elog n
26.
27. The series converges conditionally.
28. Both series converges (absolutely).
29.
f (x) = x − x5 /5 + x9 /9 − x13 /13 + . . . + C.
6
1
n2
converges
30.
log(1 + x) = x − x2 /2 + x3 /3 − x5 /5 + . . . ,
x4
1
x2
x3
log(a + x) = log a +
x−
+ 2 − 3 + . . . , |x| < a,
a
2a 3a
4a
31.
P∞
n
n=1 3n
= 3/4 and
n2
n=1 3n
P∞
R = a.
= 3/2
32.
cos x = −1 − x2 /2! + x4 /4! − x6 /6! + . . . ,
sin x = −(x − π) +
33. To show that f (x) = ex , differentiate
of ex , then evaluate the constant.
f (x)
ex
(x − π)3
(x − π)5
−
+ ....
3!
5!
using the quotient rule to show that f is a constant multiple
ex + e−x
= 1 + x2 /2! + x4 /4! + . . . .
2
34. The radius of convergence is 1, the interval is (−1, 1], and
arctan x = x − x3 /3 + x5 /5 − x7 /7 + . . .
4 terms are required to estimate π to the nearest integer, 40 terms for 1 decimal place.
35. To be updated
7