Here I’ll try to explain why computing integrals of absolute values of func-
tions requires special care. Before doing that, let’s recall one way to compute
definite integrals of absolute values of functions, that is when the integrand has
the form |f(x)|. We do this by dividing the domain up into intervals on which
f(x)≥0 and intervals on which f(x)≤0. Then adding up the results.
The example discussed in class today is:
Z3
0
|2x2−8|dx.
On the interval [0,3], we have 2x2−8≤0 when x∈[0,2] and 2x2−8≥0 when
x∈[2,3]. Alternatively, we can write
|2x2−8|=½8−2x2for x∈[0,2]
2x2−8 for x∈[2,3]
So, the definite integral is computed as
Z3
0
|2x2−8|dx =Z2
0
8−2x2dx +Z3
2
2x2−8dx
= [8x−2x3/3]2
0+ [2x3/3−8x]3
2
= (16 −16/3) + ((18 −24) −(16/3−16))
= 26 −32/3 = 46/3.
Now, if we would like to explicitly compute the antiderivative
Z|2x2−8|dx
we proceed as follows.
The generic antiderivative on the interval [0,2] is given by 8x−2x2/3 + C
while the generic antiderivative on [2,3] is given by 2x3/3−8x+C0, for constants
Cand C0. On the interval [0,3], the antiderivative is given by
Z|2x2−8|dx =½8x−2x3/3 + Cfor x∈[0,2]
2x3/3−8x+C0for x∈[2,3]
However, the constants Cand C0cannot be chosen arbitrarily: these antideriva-
tives must agree when x= 2. Therefore, we have
8(2) −2(2)3/3 + C= 2(2)3/3−8(2) + C0
16 −16/3 + C= 16/3−16 + C0
C0= 32 −32/3 + C
C0= 64/3 + C
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