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Guidelines for the test:
• Put your name or student ID number on every page.
• There are 11 problems
• The exam is closed book; calculators are not allowed.
• There is no partial credit for problem 1-3.
• For other problems,
please show all work, unless instructed otherwise. Partial credit will be given only for work shown. Print as legibly as possible - correct answers may have points taken off, if they’re illegible.
• Mark the final answer.
1. (2 pts each) f (x) is a continuous function on (−∞, ∞) and the graph of its derivative, f0(x), is shown in the figure below.
(Note: lim
x→−∞f0(x) = 0; lim
x→∞f0(x) = ∞)
Answer the following True/False questions (True ⇒ ; False ⇒ ×).
• × (1, f (1)) is an inflection point.
• f has a local maximum at x = −1.
• × f has a local minimum at x = 1.
• × f (x) has 3 critical numbers.
Sol. Since the slope of f1 0(x) at x = 1 is not zero, f00(x) 6= 0.
f2 0(−1) = 0 with f00(−1) < 0.
f3 0(1) = 0 with f00(1) < 0. The point x = 1 should be a local maximum point of f (x).
f4 0(−1) = f0(1) = f0(3) = 0, f0(0) does not exist. There are 4 critical number of f (x).
2. (2 pts each) Suppose f (x) is a continuous function, and F (x) is an antiderivative function of f (x), i.e., F0(x) = f (x). Answer the following True/False questions (True ⇒ ; False ⇒ ×).
• If f (x) is an odd function, then F (x) is an even function.
• × If f (x) is an even function, then F (x) is an odd function.
• × If f (x) is a periodic function, then F (x) is a periodic function.
• × If f (x) is monotonically increasing, then F (x) is monotonically in- creasing.
Note:
• The graph of an even function is symmetric with respect to the y-axis.
• The graph of an odd function is symmetric with respect to the origin.
• A function f is called monotonic increasing, if for all x and y such that x ≤ y one has f (x) ≤ f (y).
Sol. Let G(x) =1 Z x
0
f (t)dt be an antiderivative function of f (x), then G(−x) =
Z −x 0
f (t)dt. By letting u = −t, du = −dt, we have
G(−x) = Z −x
0
f (t)dt
= Z x
0
− f (−u)du
= Z x
0
f (u)du (since f (x) is an odd function)
= G(x).
Thus, G(x) is an even function. By the fact that the difference of F (x) and G(x) is constant, we can conclude that any antiderivative of f (x) is an even function.
Choose f (x) = x2 2 and F (x) = 13x3 + 1. We can check that f (−x) = f (x), ∀x ∈ R, but F (−x) 6= −F (x).
Choose f (x) = 1 + cos x and F (x) = x + sin x, the statement is false.3 Choose f (x) = x and F (x) =4 12x2, x ∈ R. We can check that f (x) is
strictly increasing on R, but F (x) is not.
3. (2 pts each) Answer the True/False questions (True ⇒ ; False ⇒ ×).
• 2 − cos x is an antiderivative function of sin x.
• 2 sin2 x
2 is an antiderivative function of sin x.
Sol. 1 dxd (2 − cos x) = sin x.
2 dxd 2 sin2 x2 = dxd 2 ·1−cos x2 = dxd (1 − cos x) = sin x.
4. Evaluate each of the following limits.
(a) (5 pts) lim
x→0+sin x ln x.
Sol. lim
x→0+sin x ln x = lim
x→0+
ln x
csc x. This limit is of ∞∞ form, by L’ Hospital rule,
lim
x→0+
ln x csc x
L’= lim
x→0+ d dx(ln x)
d
dx(csc x)
= lim
x→0+
1 x
− csc x cot x
= − lim
x→0+
sin x
x · tan x
= −
lim
x→0+
sin x x
lim
x→0+
tan x
= −1 · 0
= 0.
(b) (5 pts) lim
x→0+
xsin x.
Sol. Let y(x) = xsin x, we have ln y(x) = sin x ln x. By (a), we have lim
x→0+ln y(x) = lim
x→0+sin x ln x = 0. Thus, lim
x→0+xsin x = lim
x→0+y(x)
= lim
x→0+eln y(x)
= e
lim
x→0+
ln y(x)
(by the continuity of ex)
= e0
= 1.
5. Find dydx for each of the following.
(a) (5 pts) y = xsin x, x > 0.
Sol. First we have ln y(x) = ln xsin x = sin x ln x. Then,
d
dx(ln y(x)) = dxd (sin x ln x)
=⇒ y(x)1 · dxdy(x) = dxd sin x · ln x + sin x · dxd ln x
=⇒ y(x)1 dxdy(x) = cos x ln x +sin xx
=⇒ dxdy(x) = y(x) · cos x ln x + sin xx = xsin x cos x ln x + sin xx . (b) (5 pts) y = e2x
√x + 1
x2+ 2 (2x + 1)5, x > 0.
Sol. First we have ln y(x) = lnh e2x
√x+1
x2+2(2x + 1)5i
= ln e2x+ ln√
x + 1 − ln (x2+ 2)+ln (2x + 1)5 = 2x+12ln (x + 1)−ln (x2+ 2)+5 ln (2x + 1).
Then,
d
dx(ln y(x)) = dxd 2x + 12ln (x + 1) − ln (x2+ 2) + 5 ln (2x + 1)
=⇒ y(x)1 · dxdy(x) = 2 + 12 · x+11 −x21+2 · dxd (x2+ 2) + 5 · 2x+11 · dxd (2x + 1)
=⇒ y(x)1 dxdy(x) = 2 + 2x+21 −x2x2+2 +2x+110
=⇒ dxdy(x) = y(x) · 2 +2x+21 − x22x+2 + 2x+110
=h e2x
√x+1
x2+2(2x + 1)5i
2 + 2x+21 − x22x+2 +2x+110 .
6. (10 pts) Given that F (x) = Z x2
1
et2dt, for x ≥ 0.
(a) Find F0(x).
Sol. By Fundamental Theorem of Calculus, F0(x) = e(x2)2· dxdx2 = 2xex4. (b) Find (F−1)0(0).
Sol. Since F (1) = 0, we have F−1(0) = 1. Hence, (F−1)0(0) = 1
F0(F−1(0)) = 1
2F−1(0)e(F−1(0))4 = 1 2e. 7. Evaluate the given integral.
(a) (5 pts) Z
e2xsin xdx.
Sol.
Z
e2xsin xdx = 12e2xsin x − Z 1
2e2xcos xdx (by integration by parts)
= 12e2xsin x −
1
4e2xcos x + Z 1
4e2xsin xdx
(by integration by parts)
= 12e2xsin x − 14e2xcos x − 14 Z
e2xsin xdx.
We have54 Z
e2xsin xdx = 1
2e2xsin x−1
4e2xcos x+C1. Thus, Z
e2xsin xdx = 4
5
1
2e2xsin x − 1
4e2xcos x + C1
= 2
5e2xsin x − 1
5e2xcos x + C.
(b) (5 pts) Z √
ln x x dx.
Sol. Let u = ln x, du = x1dx. Then Z √
ln x x dx =
Z √
udu = 2
3u32 + C = 2
3(ln x)32 + C.
(c) (5 pts)
Z 3x
(x + 1)(x − 4)dx.
Sol.
Z 3x
(x + 1)(x − 4)dx =
Z 3 5
x + 1 +
12 5
x − 4
dx = 3
5ln |x+1|+12
5 ln |x−
4| + C.
8. (10 pts) Evaluate the definite integrals Z 4
1
e
√xdx.
Sol. Let u =√
x, du = 2√1xdx, which we have dx = 2√
xdu = 2udu. Then Z 4
1
e
√x
dx = Z 2
1
2ueudu
= 2ueu
2 1
− Z 2
1
2eudu (by integration by parts)
= 4e2− 2e −
2eu
2 1
= 4e2− 2e − (2e2− 2e)
= 2e2.
9. (a) (5 pts) Evaluate Z
cos2θdθ.
Sol.
Z
cos2θdθ =
Z 1 + cos 2θ
2 dθ = θ
2+ sin 2θ 4 + C.
(b) (5 pts) Use the trigonometric substitution to evaluate Z 1
0
√1 − x2dx.
Sol. Let x = sin θ, dx = cos θdθ. Then Z 1
0
√1 − x2dx = Z π2
0
p1 − sin2θ · cos θdθ
= Z π2
0
cos θ · cos θdθ
= Z π2
0
cos2θdθ.
By (a), we have Z 1
0
√
1 − x2dx = θ
2+sin 2θ 4
π 2
0 = π 4. 10. (5 pts) Use formulas for indefinite integrals to evaluate
Z 1
x2− 4x + 5dx.
Sol. Set a = 1, b = −4, c = 5, then b2− 4ac = 16 − 20 = −4 < 0. By using the integral formula
Z 1
ax2+ bx + cdx = 2
√4ac − b2 tan−1 2ax + b
√4ac − b2,
we have
Z 1
x2− 4x + 5dx = 2
√4tan−1 2x − 4
√4 + C = tan−1(x − 2) + C.
11. Evaluate the given integral.
(a) (5 pts) Z 1
−1
x−2dx.
Sol. Since Z 1
0
x−2dx = lim
t→0+
Z 1 t
x−2dx = lim
t→0+
−1 x
1
t = lim
t→0+
−1 + 1 t
=
∞, the integral Z 1
0
x−2dx diverges. Hence, the integral Z 1
−1
x−2dx di- verges.
(b) (5 pts) Z ∞
−∞
xdx.
Sol. Since Z ∞
0
xdx = lim
t→+∞
Z t 0
xdx = lim
t→+∞
1 2x2
t 0
= lim
t→+∞
1
2t2 = +∞, the integral
Z ∞ 0
xdx diverges. Hence, the integral Z ∞
−∞
xdx diverges.
