Name: Solutions
Calculus II
Math 116
Spring 2013
Exam 1
Exam Rules: Calculators are allowed. No Cellphones. No notes. No books. All answers should be written
on a separate sheet of paper.
(1) Evaluate the definite integral:
Z
1
(x2 + x)dx.
−1
Solution.
Z
1
1 3 1 2 1
1
1
1
x + x = (1)3 + (1)2 −
(−1)3 + (−1)2
3
2 −1
3
2
3
2
1 1
1 1
1 1 1 1
= + − − +
= + + −
3 2
3 2
3 2 3 2
2
= .
3
1
(x2 + x)dx =
−1
4
(2) Let f (x) = x2 + 1, and compute the Riemann sum of f over the interval [0, 2], choosing the representative points to be the left endpoints of the subinterval and using four subintervals of equal length.
Solution. The length of each subinterval is
2−0
2
1
(b − a)
∆x =
=
= = .
n
4
4
2
The subintervals are
1 1 3 3 0, 2 , 2 , 1 , 1, 2 , 2 , 2 .
The Riemann sum is
∆x[f (0) + f ( 21 ) + f (1) + f ( 32 )] = 12 [((0)2 + 1) + (( 21 )2 + 1) + ((1)2 + 1) + (( 32 )2 + 1))]
1
1 12 18 1 30 18
= 12 1 + 45 + 2 + 13
4 = 2 3+ 4 = 2 4 + 4 = 2 4
=
15
4
= 3.75.
4
(3) Evaluate the limit using L’Hospital’s rule:
lim
x→0
1 − cos(x)
.
x2
Solution.
cos(x)
1
1 − cos(x)
sin(x)
= lim
= .
= lim
2
x→0
x→0 2x
x→0
x
2
2
lim
4
(4) Find the area of region under the graph of f (x) = 4x3 + 2x on [0, 1].
Solution.
Z
Area =
0
1
1
(4x3 + 2x)dx = x4 + x2 = 14 + 12 − (04 + 02 ) = 2.
0
4
2
(5) Find the area of the region bounded by the graphs of f (x) = 3x2 − 4 and g(x) = 6 − 3x2 and the
vertical lines x = −1 and x = 1.
Solution.
Z
1
Z
1
2
[g(x) − f (x)]dx =
Area =
−1
2
Z
1
[(6 − 3x ) − (3x − 4)]dx =
−1
[10 − 6x2 ]dx
−1
1
= 10x − 2x3 = 10(1) − 2(1)3 − (10(−1) − 2(−1)3 ) = 10 − 2 − (−10 + 2) = 8 + 8
−1
= 16
4
(6) A car moves along a straight road in such a way that its velocity (in feet per second) at any time t
(in seconds) is given by
√
v(t) = 4 t (0 ≤ t ≤ 8)
Find the distance traveled by the car in the 8 seconds from t = 0 to t = 8.
Solution. Let D be the distance traveled by the car in the 8 seconds from t = 0 to t = 8.
8
Z 8
Z 8
Z 8 √
1
3
3
3
3
t 2 dt = 4 · 32 t 2 = 83 (8) 2 − 38 (0) 2 = 83 (8) 2 ≈ 60.34 feet.
v(t)dt =
4 tdt = 4
D=
0
0
0
0
4
(7) Find the indefinite integral.
x2
dx.
x3 − 5
Z
du
Solution. Let u = x3 − 5. Then du = 3x2 dx, so 3x
2 = dx. Then making the subsitution we have
Z
Z
Z
Z
2
2
x
x du
1
1
1
1
1
dx =
=
du =
du = ln |u| + C = ln |x3 − 5| + C.
3
2
x −5
u 3x
3u
3
u
3
3
4
(8) Evaluate the definite integral.
Z
2
2
xex dx.
1
2
Solution. Let u = x2 . Then du = 2xdx, so du
2x = dx. The new limits of integration are ua = (1) = 1
2
and ub = (2) = 4. Then making the subsitution we have
4
Z 2
Z 4
Z 4
Z
2
1
du
1 u
1 4 u
1 1
1
xex dx =
xeu
=
e du =
e du = eu = e4 − e1 = e(e3 − 1).
2x
2
2
2
2
2
2
1
1
1
1
1
4
*. (Extra Credit) Evaluate the definite integral.
Z 2
√
x x + 1dx.
0
Solution. There was a typo in this problem, the inside of the square root should have been x + 1
and not x − 1. Anyway, here’s how it’s done. Let u = x + 1. Then du = dx. So x = u − 1. The new
3
limits of integration are ua = 0 + 1 = 1 and ub = 2 + 1 = 3. Then making the subsitution we have
Z 3
Z 3
Z 3
Z 2
3
1
√
√
√
√
((u − 1) u)du =
(u u − u)du =
(u 2 − u 2 )du
x x + 1dx =
1
0
=
1
5
( 25 u 2
3
−
3 2 2 3 u )
=
5
2
2
5 (3)
1
−
3
2
2
3 (1)
=
5
2
2
5 (3)
− 23 .
1
4
limh→0 sin(h)
h
limh→0 cos(h)−1
h
**. (Extra Credit) Given that
= 1 and
= 0, use the definition of derivate
d
to show that dx (sin(x)) = cos(x).
(You’ll need to use the identity sin(A + B) = sin(A) cos(B) + cos(A) sin(B).)
Solution.
d
sin(x + h) − sin(x)
(sin(x)) = lim
h→0
dx
h
sin(x) cos(h) + cos(x) sin(h) − sin(x)
= lim
h→0
h
sin(x) cos(h) − sin(x) + cos(x) sin(h)
= lim
h→0
h
sin(x)(cos(h) − 1) + cos(x) sin(h)
= lim
h→0
h
sin(x)(cos(h) − 1)
cos(x) sin(h)
= lim
+ lim
h→0
h→0
h
h
sin(h)
cos(h) − 1
+ cos(x) lim
= sin(x) lim
h→0
h→0
h
h
= (sin(x))(0) + (cos(x))(1)
= cos(x).
4