Integration
Practice - Mat 210
I.
Basic antiderivatives. Using the
properties from your text, determine the general antiderivative below.
a)
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b)
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c)
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d)
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e)
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f)
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g)
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h)
i)
j)

k.) 
Answers:
a) F(x) = 2x + C
b) F(x) = ![]()
c)
d) ![]()
e)
f) ![]()
g)
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h)
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i)
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j)
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k) -e-x
+2x + C
II.
Initial conditions. For each function below, find
the particular antiderivative that satisfies the given initial condition.
a)
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b)
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c)
The
velocity of a particle is governed by the equation v(t) = 2t2 + t, where t is in minutes
and v(t) is in feet/minute. Find
the distance function D(t) for this particle given that after 5
minutes, the particle had traveled 30 feet.
Answers:
a)
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b)
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c)
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III.
Definite Integrals. Evaluate the following
definite integrals.
a)
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b)
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c)
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d)
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e)
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f)
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g)
The
rate of change in a town's population is given by the function f(x) = 169.147(1.07)x, where x is
years since 1980 and f(x) is people/year. What was the overall change in the town's
population between 1980 and 1990?
Answers to
definite integrals:
a)
-2
b)
174
c)
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d)
7.5
+ 3 ln 4
e)
2.6
f)
2.461
g)
2,417
people.
IV.
Area Between Two Curves. Set up the
integrals and determine the area between the two given curves between the
specified bounds. If no bounds are
given you will need to determine them (points of intersection).
a) f(x) = x2,
g(x)
= x - 2, -2 £ x £ 3
b)
f(x) = 2ex, g(x) = (1.05)x, 2 £ x £ 4
c)
Area
enclosed within the curves f(x) = x2
- 3 and g(x) = x - x2
Answers:
a)
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b)
92.16
c)
The
intersection points are x = -1 and x = 1.5. The integral would be
= 5.208