The value of $\int_{-1}^{1}|x-3| d x$ is equal to\

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Professor PC Thomas Classes · Answer

(b): Let $I=\int_{-1}^{1}|x-3| d x$\ $=\int_{-1}^{1}-(x-3) d x$\ $[\because|x-3|=-(x-3)$ for $x<3]$ $$ =\int_{-1}^{1}(-x+3) d x=\left[-\frac{x^{2}}{2}+3 xight]_{-1}^{1}=\frac{-1}{2}+3+\frac{1}{2}+3=6 $$