## Calculus: Early Transcendentals 8th Edition

$$\int^{2}_{0}|2x-1|dx=\frac{5}{2}$$
$$A=\int^{2}_{0}|2x-1|dx$$ Consider the following $$2x-1=0$$ $$x=\frac{1}{2}$$ So, from $[\frac{1}{2},2)$, $(2x-1)\gt0$, therefore $|2x-1|=2x-1$. From $(0,\frac{1}{2})$, $(2x-1)\lt0$, therefore, $|2x-1|=1-2x$ So, $A$ would become $$A=\int^{2}_{1/2}(2x-1)dx+\int^{1/2}_0(1-2x)dx$$ $$A=(\frac{2x^2}{2}-x)\Bigg]^2_{1/2}+(x-\frac{2x^2}{2})\Bigg]^{1/2}_0$$ $$A=(x^2-x)\Bigg]^2_{1/2}+(x-x^2)\Bigg]^{1/2}_0$$ $$A=[(2^2-2)-((\frac{1}{2})^2-\frac{1}{2})]+[(\frac{1}{2}-(\frac{1}{2})^2)-(0-0^2)]$$ $$A=[2+\frac{1}{4}]+[\frac{1}{4}]$$ $$A=\frac{5}{2}$$