Evaluate the following Integrals:

We have, ![]()
It can be seen that (x–1)≤0 when 0≤x≤1 and (x–1)≥0 when 1≤x≤4
= I=![]()
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= ![]()
= ![]()
= ![]()
= ![]()
= 5
Hence,
= 5
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