Calculus Quiz 13
1. (5 pts)
ˆ bx
1
a. Evaluate the limit lim 3
sin(ct2 )dt, abc 6= 0.
x→0 ax
0
b. Find r, s such that lim x−3 sin 3x + rx−2 + s = 0.
x→0
Sol.
ˆ
a. It is clear that lim ax
x→0
3
x→0
bx
sin(ct2 )dt.
= 0 = lim
By
0
L’Hospital’s rule and FTC, we have that
ˆ bx
1
b sin(b2 cx2 )
b
sin(b2 cx2 )
lim 3
sin(ct2 )dt = lim
=
lim
x→0 ax
x→0
3ax2
3a x→0
x2
0
sin b2 cx2
b3 c
b3 c
=
lim
=
3a x→0 b2 cx2
3a
sin 3x + rx + sx3
b. Let L = lim x−3 sin 3x+rx−2 +s = lim
.
x→0
x→0
x3
By L’Hospital’s rule
3 cos 3x + r + 3sx2
x→0
3x2
Since lim 3x2 = 0, the existence of L implies that
x→0
lim 3 cos 3x + r + 3sx2 = 3 + r = 0 ⇒ r = −3
L = lim
x→0
Since L = 0, so
1 − cos 3x
sin 3x − 3x
= lim
3
x→0
x
x2
3
sin 3x
9
9
= lim cos 3x =
= lim
2 x→0 x
2 x→0
2
s = − lim
x→0
1. (5 pts)
a. Evaluate the indefinite integral
ˆ
sin(ln x)dx
b. If p(x) is a polynomial with deg p = n. Show that
ˆ
n
X
p(k) (x)
ax
ax
p(x)e dx = e
(−1)k k+1 + C
a
k=0
Sol.
a.
ˆ
ˆ
y
ey sin ydy, by letting y=ln x⇒x=ey , dy= dx
= edx
y ⇒e dy=dx
x
ˆ
u=ey ,
dv=sin ydy
y
= −e cos y + ey cos ydy, by letting
du=ey dy, v=− cos y
ˆ
u=ey ,
dv=cos ydy
y
y
= −e cos y + e sin y − ey sin ydy, by letting
du=ey dy, v=sin y
ˆ
ey
ey
This shows that ey sin ydy =
sin y− cos y+C. Thus
2
2
ˆ
x
x
sin(ln x)dx = sin(ln x) − cos(ln x) + C
2
2
b. We prove the state by induction on n. For n = 0, then
p(x) = p0 for some constant p0 and
ˆ
ˆ
p0
ax
p(x)e dx = p0 eax dx = eax + C
a
Suppose that the equality hold for polynomials p with
deg p < k. Then if p(x) is a polynomial with deg p = k,
ˆ
ˆ
p(x) ax 1
u=p(x),
dv=eax dx
ax
e −
p(x)e dx =
p0 (x)eax dx, by letting
du=p0 (x)dx, v= a1 eax
a
a
Write q(x) = p0 (x), note that deg q = deg p0 = n − 1, so
ˆ
ˆ
n−1
X
q (k) (x)
0
ax
ax
ax
p (x)e dx = q(x)e dx = e
(−1)k k+1 + C
a
k=0
sin(ln x)dx =
Hence
ˆ
n−1
p(x) ax eax X
q (k) (x)
ax
p(x)e dx =
e −
(−1)k k+1 + C
a
a k=0
a
n−1
X
p(k+1) (x) (−1)k+1
+C
k+2
a
a
k=0
n
p(x) X
p(k) (x) ax
=e
+
(−1)k k+2 + C
a
a
k=1
= eax
=e
ax
p(x)
n
X
k=0
+
(−1)k
p(k) (x)
+C
ak+1
By mathematical induction hypothesis, the equality holds
for all n ∈ N.