Integral With absolute value how to solve it?

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(Its a little bit hard to see, but the limits are 0 and 2)

1 Answer
Mar 15, 2018

Please see below.

Explanation:

For #0 <= a <= 2#, we have

#abs(x(x-a)) = {(-x^2+ax,"if ",x < a),(x^2-ax,"if ",x >= a):}#

So,

#f(a) = int_0^2 abs(x(x-a)) dx#

# = int_0^a (-x^2+ax) dx + int_a^2 (x^2-ax) dx#

To find #f(a)#, evaluate the integrals.

After finding #f(a)#, find the minimum value using the usual procedure.