EXAM 1 - Math 156
NAME:
Instruction: Circle your answers and show all your work CLEARLY. Messing around may
result in losing credits, since the grader may be forced to pick the worst to grade. Solutions
with answer only and without supporting procedures will have little credits.
Z sin(x) p
1. (10 %) Find the derivative of the function
t4 + 1dt.
2
Solution: Let u = sin(x). Then by FTC and by chain rule
d
dx
Z sin(x) p
t4 + 1dt =
2
d
du
Z up
t4 + 1dt ·
2
2. (10 %) Find the average value of the function y =
du q 4
= sin (x) + 1(cos(x)).
dx
1
on the interval [0, 2].
3x + 1
Solution: The average is (as ln(1) = 0)
1
2−0
Z 2
0
dx
1 ln(3x + 1) 2 ln(7) − ln(1)
ln(7)
= ·
=
=
.
3x + 1
2
3
6
6
0
Z
3. (10 %) Use an appropriate substitution to evaluate
sin2 (x) cos(x)dx.
Solution: Set u = sin(x). Then du = cos(x)dx. (Do we have cos(x)dx in the integrand? Yes
we do. The cos(x)dx in the integrand will become du). Therefore
Z
Z
2
sin (x) cos(x)dx =
u2 du =
u3
sin3 (x)
+C =
+ C.
3
3
Z
4. (10 %) Use an appropriate substitution to evaluate
2
e(1+x ) xdx.
Solution 1: Set u = 1 + x2 . Then 12 du = xdx. (Do we have xdx in the integrand? Yes we
do. The xdx in the integrand will become 21 du). Therefore
Z
e
(1+x2 )
1
xdx =
2
2
e1+x
e du = e + C =
+ C.
2
Z
u
u
2
2
2
Solution 2: Set u = e1+x . Then by chain rule, 12 du = e1+x xdx. (Do we have e1+x xdx in
2
the integrand? Yes we do. The e1+x xdx in the integrand will become 21 du). Therefore
Z
2
e(1+x ) xdx =
1
2
2
Z
du =
u
e1+x
+C =
+ C.
2
2
Z
5. (10 %) Use an appropriate substitution to evaluate
2 sec2 (x) tan(x)
dx.
1 + sec2 (x)
Solution 1: Set u = sec(x). Then du = sec(x) tan(x)dx. (Do we have sec(x) tan(x)dx in the
integrand? Well, we can write sec2 (x) tan(x) = sec(x) · sec(x) tan(x). The sec(x) tan(x)dx in
the integrand will become du, and the leftover sec(x) becomes u). Hence
Z
2 sec2 (x) tan(x)
dx =
1 + sec2 (x)
2udu
set w = 1 + u2 , dw = 2udu
1 + u2
Z
dw
=
= ln |w| + C = ln(1 + u2 ) + C = ln(1 + sec2 (x)) + C.
w
Z
Solution 2: Set u = 1 + sec2 (x). Then du = 2 sec(x) · sec(x) tan(x)dx = 2 sec2 (x) tan(x)dx.
(Do we have 2 sec2 (x) tan(x)dx in the integrand? Yes we do. The 2 sec2 (x) tan(x)dx in the
integrand will become du). Hence
Z
2 sec2 (x) tan(x)
dx =
1 + sec2 (x)
Z
du
= ln |u| + C = ln(1 + sec2 (x)) + C.
u
Z
We use the following integration by part formula:
udv = uv −
Z
vdu.
Z
6. (10 %) Use integration by pars to evaluate
x cos(3x)dx.
Solution: Let u = x and dv = cos(3x)dx. Then v = sin(3x)/3 and du = dx. Use by-parts to
get
Z
Z
x sin(3x)
sin(3x)
x sin(3x) cos(3x)
x cos(3x)dx =
−
dx =
+
+ C.
3
3
3
9
Z
7. (10 %) Use integration by pars to evaluate
x2 ex dx.
Solution: Let u = x2 and dv = ex dx. Then v = ex and du = 2xdx and so
Z
Z
x2 ex dx = x2 ex − 2
xex dx
= x2 ex − 2xex + 2
use by-parts again with u = x and dv = ex dx
Z
ex dx
= x2 ex − 2xex + 2ex + C.
Z
8. (10 %) Evaluate
√
x3
dx
4 − x2
√
Solution 1: Let x = 2 sin(θ). Then dx = 2 cos(θ)dθ and so (using cos(θ) =
Z
x3
√
dx =
4 − x2
4−x2
):
2
(2 sin(θ))3 · 2 cos(θ)dθ
2 cos(θ)
Z
Z
= 8
= −8
sin2 (θ) sin(θ)dθ
Z
set u = cos(θ), sin2 (θ) = 1 − cos2 (θ)
cos3 (θ)
(1 − u )du = −8 cos(θ) −
3
2
3
p
= −4 4 −
x2
(4 − x2 ) 2
+
+ C.
3
!
+C
2
Solution 2: Let u = 4 − x2 . Then − du
2 = xdx, and x = 4 − u. (Do we have xdx in the
integrand? Well, we can write x3 dx = x2 · xdx. When xdx becomes 12 du, x2 becomes 4 − u and
so the integral can be converted in to an integral in u only. ) Hence
Z
x3
√
dx =
4 − x2
−1
2
=
−1
2
Z
4−u
−1
√ du =
u
2
4u
−1
+1
2
Z
(4u
1
u 2 +1
− 1
+1
2 +1
−1
2
−1
2
1
− u 2 )du
3
!
+ C = −4u
−1
2
u2
+C
+
3
3
p
= −4 4 −
Z
9. (10 %) Evaluate
x2
(4 − x2 ) 2
+
+ C.
3
ln(x)
dx.
x
Solution 1: Set u = ln(x). Then du =
Z
dx
x .
ln(x)
dx =
x
Thus
Z
udu =
(ln(x))2
+ C.
2
Solution 2: Let us try integration by parts. Our experience suggests that u = ln(x), and
dx
dv = dx
x . Thus du = x and v = ln(x). Use by-parts to get
Z
ln(x)
dx = (ln(x))2 −
x
It goes back! But there is hope. We add
Z
Hence we have the answer
R ln(x)
x
Z
ln(x)
dx.
x
Z
dx both sides to get 2
ln(x)
(ln(x))2
dx =
+ C.
x
2
ln(x)
dx = (ln(x))2 .
x
8x2 − 4x + 7
in the form of partial fractions and determine all the coefficients.
(x2 + 1)x
Solution: Write the partial fractions
10. (10 %) Write
Ax + B C
(Ax + B)x + C(x2 + 1)
8x2 − 4x + 7
=
+
=
.
(x2 + 1)x
x2 + 1
x
(x2 + 1)x
Compared the numerators
8x2 − 4x + 7 = (Ax + B)x + C(x2 + 1) = (A + C)x2 + Bx + C.
Comparing the coefficients to get C = 7, B = −4. As A + C = 8 and C = 7, we have A = 1.
Hence
8x2 − 4x + 7
x−4
7
= 2
+ .
2
(x + 1)x
x +1 x
Another way to get the coefficients: In
8x2 − 4x + 7 = (Ax + B)x + C(x2 + 1),
set x = 0, we have 7 = C. Set x = 1 with C = 7, we have 8−4+7 = A+B +14, or A+B = −3.
Set x = −1 with C = 7, we have 8 + 4 + 7 = A − B + 14, or A − B = 5. Add both A + B = −3
and A − B = 5 side by side to get 2A = 2 or A = 1. Therefore, as A + B = −3 and A = 1,
B = −4.
Grade Distribution of Exam 1:
Meaning of the scores:
at least 90 = Very good, familiar with the related materials and skillful, with minimal
computational errors. Keep on!
80-89 = good, familiar with most of the related materials, with a few computations errors.
Make an effort to do better.
70-79 = OK, not so familiar with the related materials, with relatively more computational
errors. We have room to improve. (For this quiz, not familiar with differentiation).
60-69 = Passing, We are on the borderline of failing. It indicates that we are less familiar
with the related materials and more computational errors and algebraic errors. We have lots
of room to improve.
at most 59 = We failed. We need to catch it up. It should definitely be the time for us to
see the instructor and get assistance to understand the materials and to practice MORE.
Scores
Frequency
Percentage
100
1
2.5
90-99
12
30.8
80-89
11
28.2
70-79
7
17.9
60-69
3
7.7
≤ 59
5
12.9