Area of the region bounded by the function $f(x)=\left\{\begin{array}{cc}x, & x \leq 3 \ -x+6, & x

Question

Area of the region bounded by the function $f(x)=\left\{\begin{array}{cc}x, & x \leq 3 \ -x+6, & x>3\end{array}ight.$ with the $x$-axis (in square units) in the first quadrant is\

Professor PC Thomas Classes · Answer

(b): We have, $f(x)=\left\{\begin{array}{cc}x, & x \leq 3 \ -x+6, & x>3\end{array}ight.$\ $	herefore \quad$ Required Area $=\int_{0}^{3} x d x+\int_{3}^{6}(-x+6) d x$\ $=\left.\frac{x^{2}}{2}ight|_{0} ^{3}+\left(\frac{-x^{2}}{2}+6 xight)_{3}^{6}=\frac{9}{2}+\left(\frac{-36}{2}+36+\frac{9}{2}-18ight)$\ $=9$ sq.units\