Question

Sketch the graph of $f\left(x\right)=\{\begin{array}{cc}|x-2|+2& x\le 2\\ x^2-2,& x>2\end{array}$.

Evaluate ${\int }_0^{4}f\left(x\right)dx.$

What does the value of this integral represent on the graph?

Answer 1

$f\left(x\right)=|x-2|+2$ $x⩽2$

$x^2-2$ $x>2$

$f\left(x\right)=4-x$ $x⩽2$

$x^2-2$ $x>2$

${\int }_0^{4}f\left(x\right)dx={\int }_0^2(4-x)dx+{\int }_2^4(x^2-2)dx$

${\int }_0^{4}f\left(x\right)dx=4\times (2-0)-\frac{(2^2-0)}{2}+\frac{4^3-2^3}{3}-2\times (4-2)$

$\therefore {\int }_0^{4}f\left(x\right)dx=\frac{62}{3}$