Sample Exam #3: Calculus 1
Instructions: No books, notes, or calculators are allowed.
Evaluate the following:
Z 3
x +2
1. (20 points)
dx.
x2
x3 + 2
dx =
x2
Z
Solution:
Z
1
x + 2x−2 dx = x2 − 2x−1 + C.
2
2
Z
x4 − x2 dx.
2. (20 points)
1
2
Z
4
2
x −x dx =
Solution:
1
Z
x
3. (20 points)
1 5 1 3
x − x
5
3
2
=
1
25 23
1 1
55
−
−
−
= .
5
3
5 3
15
0
2 dt .
t3
x2
Solution:
Z x
Z
0
=
2 dt
0
t3
2 dt +
x2
Z 0
x2
=
x
Z
t3
0
2 dt
t3
0
0
3
t3
2 dt + 2x
x2
"Z 2
#0
x
3
2t dt
= −
by 2nd FTC
+ 2x
3
0
(x2 )3
= −2
2x + 2x
3
by 2nd FTC and chain rule.
Calculus 1
Sample Exam #3
4. (20 points) You throw a rock upward and one second later it travelling upward
at 32 feet per second. How high will it go? Remember, acceleration due to
gravity is -32 feet per second squared.
Solution:
a(t) = −32
Z
Z
v(t) =
a(t) dt = −32 dt = −32t + C
v(1) = 32 = −32(1) + C ⇒ C = 64
∴ v(t) = −32t + 64
Z
Z
p(t) =
v(t) dt = −32t + 64 dt = −16t2 + 64t + K
p(0) = 0 = −16(02 ) + 64(0) + K
∴ p(t) = −16t2 + 64t
⇒
K=0
If t0 is time rock is highest,
0 = v(t0 ) = −32t + 64
⇒
t0 = 2
height at time t0 = 2,
p(2) = −16(22 ) + 64(2) = 64.
Z
5. (20 points) Find
√
x
dx.
x−4
Solution: Using “useful nonsense” and letting u = x − 4, one has
u
du
dx
du
dx
= x−4
= 1
= dx
= du
Calculus 1
Sample Exam #3
Substituting into
R
Z
√x
x−4
dx one has
x
√
dx =
x−4
Z
x
√ du =
u
Z
xu−1/2 du.
Noticing that u = x − 4 implies x = u + 4 one has, continuing
Z
Z
−1/2
xu
du =
(u + 4)u−1/2 du
Z
=
u1/2 + 4u−1/2 du
2 3/2
u + 8u1/2 + C
3
2
=
(x − 4)3/2 + 8(x − 4)1/2 + C.
3
=
Extra Credit
Z
6. (20 points) Find s(4) for the integral
4
x3 dx. (Here s(4) is the Riemann Sum
2
obtained by using four, equal width, inscribed rectangles.)
Solution:
3
3
1
5 1
7 1
187
3 1
+
+
s(4) = 2
+ 3
=
.
4
| {z 2} | 2{z 2} | {z 2} | 2{z 2}
box 1
box 3
box 2
box 4
3