Quiz 1 Practice Problems
Practice problems are similar, both in difficulty and in scope, to the type of
problems you will see on the quiz. Problems marked with a ? are ‘‘for your
entertainment’’ and are not essential.
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference sections
5.2 and 5.3 of the recommended textbook (or the equivalent section in your
alternative textbook/online resource) and your lecture notes.
EXPECTED SKILLS:
• Given a differentiation rule, be able to construct the associated indefinite
integration rule.
• Know how to integrate power functions (including polynomials), exponential functions, and trigonometric functions.
• Know how to simplify a ‘‘complicated integral’’ to a known form by
making an appropriate substitution of variables.
PRACTICE PROBLEMS:
For problems 1 and 2, compute the indicated derivative and state
a corresponding integration formula.
d
1
1.
dx (2x + 3)2
2.
d
[x ln x − x]
dx
1
For problems 3-16, evaluate the given indefinite integral
Z 1
2
x + x dx
3.
2
Z √
4.
x7 + e dx
Z 5.
Z
6.
Z
7.
1
+ 3x3
x3
dx
3x–2/3 + x–1/2 + 5x dx
√ 4x4/3 − 7 x dx
Z
8.
3 cos x dx
Z
9.
–7 sec2 x dx
Z 1
x
dx
10.
– +e
x
Z
11. (1 − x2 )(x3 + 4) dx
Z
x2 − 3x5
dx
x3
Z
–2 sin x
dx
cos2 x
12.
13.
Z
14.
(6 cos x + 9 csc2 x) dx
Z
(sin x − 3 sec x tan x) dx
15.
Z
16.
2x dx
2
Z
17. Consider
cot2 x dx
(a) Using the fact that sin2 x + cos2 x = 1, derive the identity cot2 x =
csc2 x − 1.
(b) Use the identity that you derived in part (a) to evaluate the
original integral.
For problems 18 and 19, find a function y = y(x) which satisfies the
given initial value problem.

1
dy


= 2

 dx
9x
18.



 y(1) = 1
2

 dy = –2ex
dx
19.
 y(0) = –5
20. A ball is thrown straight up in the air from an initial height of s0
feet above the ground with an initial speed of v0 ft/sec. Then s(t)
gives the height (in feet) above the ground at time t, v(t) = s0 (t) gives
the velocity (in ft/sec) of the ball at time t, and a(t) = s00 (t) gives the
acceleration (in ft/sec2 ) of the ball at time t. Assuming that acceleration
is a constant –g ft/sec2 , determine v(t) and s(t).
For problems 21-39, evaluate the given indefinite integral.
Z
21. 3x2 (x3 + 3)3 dx
Z
22.
Z
23.
5
dx
5x + 3
2x cos(x2 ) dx
3
Z
24.
4x(x2 + 6)2 dx
Z
25.
sec(4x) tan(4x) dx
Z
26.
Z
27.
(3x − 5)9 dx
e–2x dx
Z
sin x cos x
dx
1 + sin2 x
Z
x
x
cot
dx
29.
csc
2
2
Z
√
30. –3x3 1 − x4 dx
28.
Z 31.
Z
1
dx
2 + 4x2
Z
4x
dx
(3 + x2 )2
Z
√
x2 4 − x dx
Z
e3/x
dx
x2
Z
ex
dx
e2x + 1
32.
33.
34.
35.
36.
Z
37.
38.
√
e x 1
√ − cos 4x
x 4
dx
(sin 4x)(cos 4x)2/3 dx
Z 3
csc2 (3x) tan2 (3x) + x2 ex
dx
4
Z
39.
1
dx
x ln x
Z
? Determine
f (x) dx where
f (x) =
Z
? Compute



 x + 1,
x<0
0,
0 ≤ x ≤ 20


 5 − x2 , x > 20.
x + sin x − cos x − 1
dx.
x + ex + sin x
(Taken from a Romanian college entrance exam.)
5
SOLUTIONS
1. By chain rule,
1
–2
d
–4
=
·2=
,
2
3
dx (2x + 3)
(2x + 3)
(2x + 3)3
which gives us
Z
1
–4
dx =
+ C.
3
(2x + 3)
(2x + 3)2
2. By the product rule,
d
[x ln x − x] = (1)(ln x) + (x)
dx
1
− 1 = ln x + 1 − 1 = ln x,
x
which gives us
Z
ln x dx = x ln x − x + C.
1
1 x2 x3
1
1
2
3.
x + x dx = ·
+
+ C = x2 + x3 + C
2
2 2
3
4
3
Z √
Z
2
7
x + e dx =
x7/2 + e dx = x9/2 + ex + C
4.
9
Z Z
1
1
3
5.
+ 3x3 dx = (x–3 + 3x3 ) dx = – x–2 + x4 + C
3
x
2
4
Z 6
Z
5
3x–2/3 + x–1/2 + 5x dx = 9x1/3 + 2x1/2 + x2 + C
2
Z
Z
√ 7x3/2
4x7/3
4/3
4x − 7 x dx =
7.
−
+C =
4x4/3 − 7x1/2 dx =
7/3
3/2
12 7/3 14 3/2
x − x +C
7
3
Z
8. 3 cos x dx = 3 sin x + C
6.
Z
9.
–7 sec2 x dx = –7 tan x + C
Z 1
x
10.
– +e
dx = – ln |x| + ex + C
x
Z
Z
1
1
2
3
11. (1 − x )(x + 4) dx FOIL
(x3 + 4 − x5 − 4x2 ) dx = x4 + 4x − x6 −
=
4
6
4 3
x +C
3
Z Z 2
1
x − 3x5
2
dx =
− 3x dx = ln |x| − x3 + C
12.
x3
x
Z
Z
Z
–2 sin x
–2
sin x
dx =
13.
·
dx = –2 sec x tan x dx = –2 sec x + C
cos2 x
cos x cos x
Z
14.
(6 cos x + 9 csc2 x) dx = 6 sin x − 9 cot x + C
Z
(sin x − 3 sec x tan x) dx = – cos x − 3 sec x + C
15.
Z
16.
x
2 dx =
Z
e
ln 2 x
Z
dx =
17. (a) sin2 x + cos2 x = 1 ⇒
e(ln 2)x dx =
2x
1 (ln 2)x
e
+C =
+C
ln 2
ln 2
sin2 x cos2 x
1
=
⇒ 1 + cot2 x = csc2 ⇒
2 +
2
sin x sin x
sin2 x
cot2 x = csc2 x − 1
Z
Z
2
(b)
cot x dx = (csc2 x − 1) dx = – cotx − x + C
7
Z
18. We have y =
dy
dx =
dx
Z
1
1
1
dx = – + C. Since y(1) = ,
2
9x
9x
2
1
1
1 1
11
=–
+C ⇒C = + = ,
2
9(1)
2 9
18
and so y(x) = –
Z
19. We have y =
1
11
+ .
9x 18
dy
dx =
dx
Z
–2ex dx = –2ex + C. Since y(0) = –5,
–5 = –2e0 + C = –2 + C ⇒ C = 2 − 5 = –3,
and so y(x) = –2ex − 3.
20. Let us first find v(t). Since a(t) = v 0 (t), the given data yields the
following initial value problem:

 dv = –g
dt
 v(0) = v .
0
Z
Hence v(t) =
0
Z
v (t) dt =
–g dt = –gt + C and
v0 = –g(0) + C = C
so that v(t) = –gt + v0 . We can now find s(t) by way of the other
available initial value problem:

 ds = –gt + v
0
dt
 s(0) = s .
0
8
Z
Z
0
Therefore s(t) =
s (t) dt =
1
(–gt + v0 ) dt = – gt2 + v0 t + C and
2
1
s0 = – g(0)2 + v0 (0) + C = C
2
1
so that s(t) = – gt2 + v0 t + s0 . (Observe that these are the formulas
2
describing free fall that one learns in physics.)
21. Let u = x3 + 3 so that du = 3x2 dx. Then
Z
2
3
Z
3
3x (x + 3) dx =
1
1
u3 du = u4 + C = (x3 + 3)4 + C.
4
4
22. Let u = 5x + 3 so that du = 5 dx. Then
Z
5
dx =
5x + 3
Z
1
du = ln|u| + C = ln|5x + 3| + C.
u
23. Let u = x2 so that du = 2x dx. Then
Z
2
Z
2x cos(x ) dx =
cos u du = sin u + C = sin(x2 ) + C.
24. Let u = x2 + 6 so that du = 2x dx ⇒ 2du = 4x dx. Then
Z
2
2
4x(x + 6) dx = 2
Z
2
2
u2 du = u3 + C = (x2 + 6)3 + C.
3
3
9
1
25. Let u = 4x so that du = 4 dx ⇒ du = dx. Then
4
Z
1
sec(4x) tan(4x) dx =
4
Z
sec u tan u dx =
1
1
sec u+C = sec(4x)+C.
4
4
1
26. Let u = 3x − 5 so that du = 3 dx ⇒ du = dx. Then
3
Z
1
(3x − 5) dx =
3
9
Z
u9 du =
1 1 10
1
· u + C = (3x − 5)10 + C.
3 10
30
1
27. Let u = –2x so that du = –2 dx ⇒ – du = dx. Then
2
Z
Z
1
1
1
–2x
e dx = –
eu du = – eu + C = – e–2x + C.
2
2
2
1
28. Let u = 1 + sin2 x so that du = 2 sin x cos x dx ⇒ du = sin x cos x dx.
2
Then
Z
sin x cos x
1
dx =
2
2
1 + sin x
Z
1
1
1
1
du = ln|u|+C = ln|1+sin2 x|+C = ln(1+sin2 x)+C.
u
2
2
2
(Note that because 1 + sin2 x > 0 for all real x, the absolute value is
superfluous and can be removed.)
29. Let u =
Z
csc
x
1
so that du = dx ⇒ 2du = dx. Then
2
2
x
2
cot
x
2
Z
dx = 2
csc u cot u du = –2 csc u+C = –2 csc
10
x
+C.
2
3
30. Let u = 1 − x4 so that du = –4x3 dx ⇒ du = –3x3 dx. Then
4
Z
Z
√
√
3
3 2
1
–3x 1 − x4 dx =
u du = · u3/2 + C = (1 − x4 )3/2 + C.
4
4 3
2
3
√
Z
e x
1
√
31. First split the integral in two:
cos 4x dx. The second
dx −
4
x
√
integral is pretty easy. For the first one, let u = x so that du =
1
1
√ dx ⇒ 2du = √ dx. Then
2 x
x
Z
Z
√
e x
√ dx =
x
Z
√
e
x
1
· √ dx = 2
x
Z
eu du = 2eu + C = 2e
√
x
+ C.
Combining this with the other integral, we see that
Z √
e x 1
√ − cos 4x
x 4
√
dx = 2e
x
−
1
sin 4x + C.
16
32. One may suspect that the answer will involve the inverse tangent
function, but instead of a ‘‘1 + something squared’’ we have a ‘‘2 +
something squared.’’ That is fine. To mitigate that, let us factor out a
2 from the denominator of the integrand so that
Z
1
dx =
2 + 4x2
Z
1
1
1
·
dx =
2
2 1 + 2x
2
It now makes sense to make u =
dx. Then
11
√
Z
1+
2x so that du =
1
√
√
2x
2 dx.
1
2 dx ⇒ √ du =
2
1
2
Z
1
√
1+
1 1
2 dx = · √
2 2
2x
Z
√ 1
1
1
–1
–1
2x +C.
du = √ tan u+C = √ tan
1 + u2
2 2
2 2
33. Let u = 3 + x2 so that du = 2x dx ⇒ 2du = 4x dx. Then
Z
4x
dx = 2
(3 + x2 )2
Z
1
–2
1
du = –2 · + C =
+ C.
2
u
u
3 + x2
34. This is very tricky, but it is best to let u = 4 − x so that x = 4 − u ⇒
dx = –du. So
Z
x
2
√
Z
4 − x dx = –
Z
=–
Z
=–
√
(4 − u)2 u du
(16 − 8u + u2 )u1/2 du
16u1/2 − 8u3/2 + u5/2 du
2 5/2 2 7/2
2 3/2
+C
= – 16 · u − 8 · u + u
3
5
7
32
16
2
= – (4 − x)3/2 + (4 − x)5/2 − (4 − x)7/2 + C.
3
5
7
35. Let u =
Z
3
3
1
1
so that du = – 2 dx ⇒ – du = 2 dx. Then
x
x
3
x
e3/x
dx =
x2
Z
3/x
e
1
1
· 2 dx = –
x
3
12
Z
1
1
eu du = – eu + C = – e3/x + C.
3
3
36. Let u = ex so that du = ex dx. Then
Z
ex
dx =
e2x + 1
Z
du
= tan–1 u + C = tan–1 (ex ) + C.
+1
u2
1
37. Let u = cos 4x so that du = –4 sin 4x dx ⇒ – du = sin 4x dx. Then
4
Z
2/3
(sin 4x)(cos 4x)
1
dx = –
4
Z
3
1 3
u2/3 du = – · u5/3 +C = – (cos 4x)5/3 +C.
4 5
20
Z
2
38. First split the integral in two:
2
csc (3x) tan (3x) dx+
Z
3
x2 ex dx. Note
that
Z
2
sin2 (3x)
1
·
dx
sin2 (3x) cos2 (3x)
Z
1
=
dx
2
cos (3x)
Z
1
= sec2 (3x) dx = tan(3x) + C.
3
Z
2
csc (3x) tan (3x) dx =
For the second term, let u = x3 so that du = 3x2 dx ⇒
Then
Z
1
x e dx =
3
2 x3
Z
1
du = x2 dx.
3
1
1 3
eu du = eu + C = ex + C.
3
3
Hence
13
Z 2
2
csc (3x) tan (3x) + x e
39. Let u = ln x so that du =
Z
2 x3
1
dx =
x ln x
Z
dx =
1
1 3
tan(3x) + ex + C.
3
3
1
dx. Then
x
1 1
· dx =
ln x x
Z
14
1
du = ln |u| + C = ln |ln x| + C.
u