Question

By using the properties of definite integrals, evaluate the integral: $\int_{0}^{4}|x-1|dx$

Answer 1

I=$\int_{0}^{4}|x-1|dx$

It can be seen that,(x−1)≤0 when 0≤x≤1and(x−1)≥0 when1≤x≤4

I=$\int_{0}^{1}|x-1|dx$+$\int_{1}^{4}|x-1|dx$ $\,\,\,\,\,\ \because\int_{a}^{b}f(x)=\int_{0}^{a}f(x)+\int_{b}^{c}f(x)$

=$\int_{0}^{1}-(x-1)dx+\int_{0}^{4}(x-1)dx$

$=1- (\frac{1}{2}) +\frac{(4^2)} {2} -4- (\frac{1}{2}) +1$

=5