Practice with Definite Integrals
Use antidifferentiation to compute the following definite integrals. Check your
work using the geometric definition of the definite integral, graphing and esti­
mation.
� 2
a)
x2 dx
0
�
e
b)
1
�
1
dx
x
0
c)
sin x dx
−π/4
Solution
� 2
a)
x2 dx
0
The antiderivative of x2 is 13 x3 (plus a constant), so:
�
�2
1 3 ��
x
3 �0
1 3 1 3
·2 − ·0
3
3
8
.
3
2
2
x dx
=
0
=
=
6
5
4
3
2
1
0
−2
−1
0
1
1
2
Geometrically, the area under the curve appears to be between 2 and 4 so
our answer seems to be correct.
�
e
b)
1
1
dx
x
The antiderivative of
1
x
is ln |x|.
� e
1
dx
x
1
e
=
[ln |x|]1
=
ln e − ln 1
=
=
1−0
1
3
2
1
0
0
1
2
3
The area under the curve does appear to be between 1 and 1.5.
�
0
c)
sin x dx
−π/4
The antiderivative of sin x is − cos x.
� 0
0
sin x dx = − cos x|−π/4
−π/4
=
=
− cos(0) − (− cos(−π/4))
1
−1 + √
2
2
√
1− 2
=
2
≈ −0.2
0.5
0
−1.2 −0.8 −0.4
0
0.4
0.8
−0.5
We’re computing the area of a region that lies below the x-axis, so we expect
the answer to be negative. The estimate −0.2 seems very small but is in fact
close to the correct value, as we see when we compare the shaded region to
a rectangle with side lengths 0.5 and 0.4.
3
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18.01SC Single Variable Calculus��
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