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Question

By using the properties of definite integrals, evaluate the integral 40|x1|dx

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Solution

I=40|x1|dx
It can be seen that, (x1)0 when 0x1 and (x1)0 when 1x4
I=10|x1|dx+41|x1|dx,(baf(x)=caf(x)+bcf(x))
=10(x1)dx+40(x1)dx
=[xx22]10+[x22x]41
=112+(4)22412+1
=112+8412+1=5

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