Calculus 1, Quiz 5 Instructor: Frank Liou
Exam Time: 20 min
Remark. No partial credit if the answer is wrong. Check your solution carefully before you turn in your answer sheet.
(1) (5 Points) Calculate d dx
Z tan x sin x
p1 + t2dt.
Solution:
d dx
Z tan x sin x
p1 + t2dt =p
1 + tan2· sec2x −p
1 + sin2x cos x.
(2) (5 Points) Let f (x) = x + x2+ · · · + xn, x ∈ R. Then f0(x) = 1 + 2x + · · · + nxn−1. (a) (3 Points) Evaluate
n
X
i=1
i
3i−1 using the formula for f0(x).
(b) (2 Points) Find
∞
X
n=1
n
3n−1 using (a).
Solution:
Let g(x) = x + · · · + xn, x ∈ R. Since
f (x) = x + x + · · · + xn= x(1 − xn)
1 − x = x − xn+1 1 − x . Hence
f0(x) = (1 − (n + 1)xn)(1 − x) + (x − xn+1)
(1 − x)2 = 1 − (n + 1)xn+ nxn+1 (1 − x)2 . The n-th partial sum of
∞
X
n=1
n
3n−1 is given by
n
X
i=1
i 3i−1 = 9
4
1 −n + 1 3n + n
3n+1
. By taking n → ∞, we obtain
∞
X
n=1
n 3n−1 = 9
4.
(3) (5 Points) The change of variable formula is given by Z g(d)
g(c)
f (x)dx = Z d
c
f (g(u))g0(u)du
1
2
where g ∈ C1[a, b] and increasing. Evaluate Z
√ 3 2
√ 2 2
p1 − x2dx
using the change of variable x = sin u. Here x = sin u is increasing on π
4 ≤ u ≤ π 3. Solution:
Z
√ 3 2
√ 2 2
p1 − x2dx = Z π
3
π 4
p
1 − sin2u · cos udu
= Z π
3
π 4
cos2udu
= 1 2
Z π
3
π 4
(1 + cos 2u)du
= 1 2
π 3 − π
4
+1
4
sin 2 · π
3 − sin 2 ·π 4
= π 24 +
√ 3 − 2
8 .
(4) (5 Points) The integration by part formula is given by Z b
a
f (x)g0(x)dx = f (x)g(x)|x=bx=a− Z b
a
g(x)f0(x)dx.
Evaluate
Z π 0
x cos nxdx
using integration by part formula. Here n is a natural number.
Solution:
Z π 0
x cos nxdx = x sin nx n
π 0
− 1 n
Z π 0
sin nxdx
= 1
n2(cos nπ − 1)
= (−1)n− 1 n2 .
