Quiz 12
1 t2 − 1
1. (a) Evaluate the integral ∫
dt.
0 t4 − 1
Solution:
∫
1
√
3
0
1
1
1
√
√
√
t2 − 1
t2 −
1
1
π
3
3
3
= .
dt
=
dt
=
dt
=
arctan
t∣
∫
∫
2
4
2
2
0
t −1
(t − 1) (t + 1)
t +1
6
0
0
1 sin x
dx = 0.
(b) Explain why ∫
−1 1 + x2
sin x
Solution: The function f (x) = 1+x
2 is an odd function on the interval [−1, 1], f (−x) = −f (x).
2. What is wrong with the equation?
∫
Solution: The integrand f (x) =
not applicable.
1
cos2 x
π
0
π
1
dx = tan x∣ = 0.
2
0
cos x
is discontinuous at x =
π
,
2
hence the Evaluation Theorem is
3. Use the Fundamental Theorem of Calculus to find the derivative of y = ∫
Solution: First we rewrite
y(x) = ∫
= −∫
cos x
sin x
sin x
0
(1 + v 2 )10 dv = ∫
(1 + v 2 )10 dv + ∫
0
sin x
cos x
0
(1 + v 2 )10 dv + ∫
cos x
0
cos x
sin x
(1 + v 2 )10 dv.
(1 + v 2 )10 dv
(1 + v 2 )10 dv
and then apply the Fundamental Theorem of Calculus
y ′ (x) = −(1 + sin2 x)10 ⋅ cos x + (1 + cos2 x)10 ⋅ (− sin x)
4. Find the average value of the function f (x) =
1
x
on the interval [1, 4].
Solution:
fave =
4 1
4
2
1
1
dx = ln ∣x∣∣ = ln 2.
∫
1
4−1 1 x
3
3
5. Use the Substitution Rule to evaluate:
(a) the indefinite integral ∫ 5t sin(5t )dt
Solution:
t
t
======
∫ 5 sin(5 )dtdu=5
t ln 5 dt
u(t)=5t
1
1
cos(5t ) + C.
∫ sin u du = −
ln 5
ln 5
1 ez + 1
(b) the definite integral ∫
dz
0 ez + z
Solution:
∫
z=1
z=0
u=e+1 1
u=e+1
ez + 1
u(z)=ez +z
dz
======
du = ln ∣u∣ ∣
= ln(e + 1)
∫
z
z
u=1
e + z du=(e +1) dz u=1
u