The integral _0^2||x-1|-x|,dx is equal to _______.

Question

The integral $\displaystyle\int_0^2\big||x-1|-x\big|\,dx$ is equal to \_\_\_\_\_\_\_.

Answer: 1.50

Accepted Answer

For $x\in[0,1]$, $|x-1|-x=1-2x$; for $x\in[1,2]$, $|x-1|-x=-1$.
$$I=\int_0^1|1-2x|\,dx+\int_1^2|-1|\,dx=\int_0^{1/2}(1-2x)dx+\int_{1/2}^1(2x-1)dx+\int_1^2dx$$

$$=\left[x-x^2\right]_0^{1/2}+\left[x^2-x\right]_{1/2}^1+\left[x\right]_1^2=\frac14+\frac14+1=\frac32=1.50$$

Solutions · 1