Here’s a list of practice exercises. There’s a hint
for each one as well as an answer with intermediate
steps.
Practice Integration
Math 120 Calculus I
Z
1.
(x4 − x3 + x2 ) dx. Hint. Answer.
D Joyce, Fall 2013
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(5t8 − 2t4 + t + 3) dt. Hint. Answer.
This first set of indefinite integrals, that is, an- 2.
tiderivatives, only depends on a few principles of
Z
integration, the first being that integration is inverse to differentiation. Besides that, a few rules 3.
(7u3/2 + 2u1/2 ) du. Hint. Answer.
can be identified: a constant rule, a power rule,
linearity, and a limited few rules for trigonometric,
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logarithmic, and exponential functions.
4.
(3x−2 − 4x−3 ) dx. Hint. Answer.
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k dx = kx + C, where k is a constant
Z
3
5.
dx. Hint. Answer.
Z
x
1
xn dx =
xn+1 + C, if n 6= −1
n+1
Z 7
4
Z
+
6.
dt. Hint. Answer.
1
3t2 2t
dx = ln |x| + C
x
Z
Z
Z 3
√
kf (x) dx = k f (x) dx
7.
dy. Hint. Answer.
5 y−√
y
Z
Z
Z
(f (x) ± g(x)) dx = f (x) dx ± g(x) dx
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3x2 + 4x + 1
Z
8.
dx. Hint. Answer.
2x
sin x dx = − cos x + C
Z
Z
9.
(2 sin θ + 3 cos θ) dθ. Hint. Answer.
cos x dx = sin x + C
Z
ex dx = ex + C
Z
10.
(5ex − e) dx. Hint. Answer.
Z
1
dx = arctan x + C
Z
1 + x2
4
Z
11.
dt. Hint. Answer.
2
1
1
+
t
√
dx = arcsin x + C
1 − x2
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We’ll add more rules later, but there are plenty here
12.
(ex+3 + ex−3 ) dx. Hint. Answer.
to get acquainted with.
Z
13.
1
√
7
du. Hint. Answer.
1 − u2
Z 14.
Z
15.
Z
16.
17.
1
r − 2r +
r
2
Integrating polynomials is fairly easy, and you’ll
get the hang of it after doing just a couple of them.
Answer.
dr. Hint. Answer.
4 sin x
dx. Hint. Answer.
3 tan x
Z
3. Hint.
You can use the power rule for other powers besides integers. For instance,
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u3/2 du = 52 u5/2 + C
(7 cos x + 4ex ) dx. Hint. Answer.
Z √
3
(7u3/2 + 2u1/2 ) du.
7v dv. Hint. Answer.
Answer.
Z
18.
Z
19.
Z
20.
4
√ dt. Hint. Answer.
5t
Z
4. Hint.
(3x−2 − 4x−3 ) dx
You can even use the power rule for negative exponents (except −1). For example,
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x−3 dx = − 12 x−2 + C
1
dx. Hint. Answer.
3x2 + 3
√
x4 − 6x3 + ex x
√
dx. Hint. Answer.
x
Answer.
Z
3
dx
x
This is 3x−1 and the general power rule doesn’t
apply. But you can use
Z
1
dx = ln |x| + C.
x
5. Hint.
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1. Hint.
(x4 − x3 + x2 ) dx.
Integrate each term using the power rule,
Z
1
xn dx =
xn+1 + C.
n+1
Answer.
So to integrate xn , increase the power by 1, then
6. Hint.
divide by the new power. Answer.
Z
2. Hint.
4
7
+
dt
3t2 2t
Treat the first term as 34 t−2 and the second term
as 72 t−1 . Answer.
(5t8 − 2t4 + t + 3) dt.
Z Z Remember that the integral of a constant is the
3
√
5 y−√
dy
constant times the integral. Another way to say 7. Hint.
y
that is that you can pass a constant through the
It’s usually easier to turn those square roots into
integral sign. For instance,
1
fractional powers. So, for instance, √ is y −1/2 .
Z
Z
y
5t8 dt = 5 t8 dt
Answer.
2
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3x2 + 4x + 1
16. Hint.
(7 cos x + 4ex ) dx
dx
8. Hint.
2x
Just more practice with trig and exponential
Use some algebra to simplify the integrand, that
functions. Answer.
is, divide by 2x before integrating. Answer.
Z √
Z
3
17. Hint.
7v dv
9. Hint.
(2 sin θ + 3 cos θ) dθ
√
√
√
3
3
3
You can write
7v
as
7
v. And remember
Getting the ± signs right when integrating sines
√
1/3
3
you can write v as v . Answer.
and cosines takes practice. Answer.
Z
Z
4
√ dt
18. Hint.
10. Hint.
(5ex − e) dx
5t
Use algebra to write this in √
a form that’s easier to
Just as the derivative of ex is ex , so the integral
x
x
t is t−1/2 . Answer.
integrate.
Remember
that
1/
of e is e . Note that the −e in the integrand is a
constant. Answer.
Z
1
dx
19. Hint.
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3x2 + 3
4
11. Hint.
dt
You can factor out a 3 from the denominator to
1 + t2
Remember that the derivative of arctan t is put it in a form you can integrate. Answer.
1
. Answer.
√
Z 4
1 + t2
x − 6x3 + ex x
√
20. Hint.
dx
x
Z
√
Divide through by x before integrating. Alter12. Hint.
(ex+3 + ex−3 ) dx
natively, write the integrand as
When working with exponential functions, rex−1/2 (x4 − 6x3 + ex x1/2 )
member to use the various rules of exponentiation. Here, the rules to use are ea+b = ea eb and
and multiply. Answer.
ea−b = ea /eb . Answer.
Z
Z
7
du
1 − u2
Remember that the derivative of arcsin u is
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1
√
Answer.
1. Answer.
(x4 − x3 + x2 ) dx.
1 − u2
13. Hint.
√
Z The integral is 15 x5 − 14 x4 + 13 x3 + C.
1
Whenever you’re working with indefinite inte14. Hint.
r2 − 2r +
dr
r
grals like this, be sure to write the +C. It signifies
Use the power rule, but don’t forget the integral that you can add any constant to the antiderivative
of 1/r is ln |r| + C. Answer.
F (x) to get another one, F (x) + C.
When you’re working
Z
Z bwith definite integrals with
4 sin x
15. Hint.
dx
limits of integration,
, the constant isn’t needed
3 tan x
a
You’ll need to use trig identities to simplify this. since you’ll be evaluating an antiderivative F (x) at
Answer.
b and a to get a numerical answer F (b) − F (a).
3
Z
8
Z
4
(5t − 2t + t + 3) dt.
2. Answer.
10. Answer.
(5ex − e) dx
That equals 5ex − ex + C.
The integral is 95 t9 − 52 t5 + 21 t2 + 3t + C.
Z
Z
3. Answer.
(7u
3/2
+ 2u
This integral evaluates as
Z
4. Answer.
1/2
4
dt.
1 + t2
That evaluates as 4 arctan t + C. Some people
prefer to write arctan t as tan−1 t.
11. Answer.
) du.
14 5/2
u
5
+ 43 u3/2 + C.
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(3x−2 − 4x−3 ) dx.
12. Answer.
(ex+3 + ex−3 ) dx.
The integrand is its own antiderivative, that is,
That equals −3x−1 +2x−2 +C. If you prefer, you
the integral is equal to
3
2
could write the answer as − + 2 + C
x x
ex+3 + ex−3 + C.
Z
3
If you write the integrand as ex e3 + ex /e3 , and note
dx
5. Answer.
that e3 is just a constant, you can see that it’s its
x
That’s 3 ln |x|+C. The reason the absolute value own antiderivative.
sign is there is that when x is negative, the derivaZ
7
tive of ln |x| is 1/x, so by putting in the absolute 13. Answer.
√
du.
1 − u2
value sign, you’re covering that case, too.
The integral equals 7 arcsin u.
Z Z 7
4
1
+
6. Answer.
dt.
2
2
14. Answer.
r − 2r +
dr.
3t
2t
r
The integral of 43 t−2 + 72 t−1 is − 34 t−1 + 27 ln |t| + C.
The integral evaluates as
Z 7. Answer.
3
√
5 y−√
y
1 3
r
3
dy.
− r2 + ln |r| + C.
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4 sin x
15. Answer.
dx
The integral of 5y −3y
is
3 tan x
10 √
You could write that as 3 y y − 6 y + C if you
The integrand simplifies to 43 cos x. Therefore the
prefer.
integral is 34 sin x + C.
Z
Z
3x2 + 4x + 1
8. Answer.
dx.
16. Answer.
(7 cos x + 4ex ) dx.
2x
The integral of 2x + 2 + 21 x−1 is
That’s 7 sin x + 4ex + C.
1/2
−1/2
10 3/2
y −6y 1/2 +C.
3 √
Z √
3
17. Answer.
7v dv.
1
x + 2x + ln |x| + C.
2
2
Since you can rewrite the integrand as
therefore its integral is
√ 4/3
3 3
7 v + C.
4
Z
9. Answer.
(2 sin θ + 3 cos θ) dθ.
That’s equal to −2 cos θ + 3 sin θ + C.
4
√
3
7 v 1/3 ,
Z
4
√ dt.
5t
4 −1/2
8
The integral of √ t
is equal to √ t1/2 + C.
5
5
p
You could also write that as 8 t/5 + C.
18. Answer.
Z
1
dx
+3
This integral equals 13 arctan x + C.
19. Answer.
3x2
√
x4 − 6x3 + ex x
√
20. Answer.
dx.
x
The integral can be rewritten as
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(x7/2 − 6x5/2 + ex ) dx
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which equals 92 x9/2 −
12 7/2
x
7
+ ex + C.
Math 120 Home Page at
http://math.clarku.edu/~djoyce/ma120/
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