Download Calculus Quiz 10

Calculus Quiz 10
1. (5 pts)
ˆ ˆx
t2
√
1 + u4
dudt. Find F 00 (2).
√
u
1
t
b. Find a function f and a number a such that
ˆ x
√
f (t)
6+
x, for all x > 0
dt
=
2
t2
a
a. Let F (x) =
Sol.
a. By FTC, we know that
ˆ x2 √
1 + u4
0
F (x) = √
du,
u
x
and
p
p
√
1 + (x2 )4
1 + ( x)4
1
√
F (x) =
· √
· 2x −
2
x
x
2 x
√
√
2 1 + x8
1 + x2
=
−
x
2x
√
√
5
Hence F 00 (2) = 257 −
4
b. Differentiate the equality with respect to x on both side,
then by FTC, we get
00
3
f (x)
1
= √ ⇒ f (x) = x 2
2
x
x
Substituting x = a into the equality, we have that
√
6=2 a⇒a=9
2. (5 pts)
a. If f is continuous on [0, π], show that
ˆ π
ˆ
π π
xf (sin x)dx =
f (sin x)dx
2 0
0
[Hint: Taking u = π − x]
b. Use a. to evaluate the integral
ˆ π
x sin2 x cos4 xdx
0
Sol.
a.
ˆ π
ˆ
0
(π − u)f sin(π − u) (−du), by letting u=1−x⇒du=−dx
0
ˆπ π
ˆ π
ˆ π
=
(π − u)f (sin u)du = π
f (sin u)du −
uf (sin u)du
0
0
0
ˆ π
ˆ π
xf (sin x)dx
f (sin u)du −
=π
0
0
ˆ π
ˆ π
Thus we get 2 xf (sin x)dx = π
f (sin x)dx and therexf (sin x)dx =
fore
0
ˆ
0
ˆ
π π
xf (sin x)dx =
f (sin x)dx
2 0
0
b. By a., we have that
ˆ
ˆ π
ˆ π
π π 2
2
4
2
2
2
sin x(1 − sin2 x)2 dx
x sin x cos xdx =
x sin x(1 − sin x) dx =
2
0
0
ˆ
ˆ π 0
π π 2
π
=
sin x cos4 xdx =
(sin x cos x)2 cos2 xdx
2 0
2 0
ˆ
ˆ
π π 2 1 + cos 2x π π 2
=
sin 2x
sin 2x(1 + cos 2x)dx
dx =
8 0
2
16 0
ˆ
ˆ
π π 2
π π 2
sin 2xdx +
sin 2x cos 2xdx
=
16 0
16 0
ˆ
ˆ
π π 1 − cos 4x
π 0 2
=
dx +
u du, by letting u=sin 2x⇒du=2 cos 2xdx
16 0
2
32 0
ˆ π
ˆ π
π
π
=
dx −
cos 4xdx
32 0
32 0
ˆ 4π
π2
π
=
cos vdv, by letting v=4x⇒dv=4dx
−
32 128 0
v=4π π 2
π2
π
=
−
sin v =
32 128
32
v=0
π