## Calculus 8th Edition

$\displaystyle \frac{5}{2}$
$\displaystyle \int_2^5|x-3|dx$ Since the graph of $y=x-3$ is negative for $x<3$, we must split this integral into two parts to assure that we are getting the absolute value of $y=x-3$ $\displaystyle -\int_2^3(x-3)dx+\int_3^5(x-3)dx$ The reason why the first integral is negative is because we need to subtract out the negative area, thus we are taking the "absolute value" since we are adding the area both under the $x$-axis(from $x=2$ to $x=3$) and above (from $x=3$ to $x=5$). Now we evaluate: $\displaystyle -(\frac{x^2}{2}-3x)|_2^3+(\frac{x^2}{2}-3x)|_3^5$ $\displaystyle -([\frac{(3)^2}{2}-3(3)]-[\frac{(2)^2}{2}-3(2)])+([\frac{(5)^2}{2}-3(5)]-[\frac{(3)^2}{2}-3(3)])$ $\displaystyle -([\frac{9}{2}-9]-[\frac{4}{2}-6])+([\frac{25}{2}-15]-[\frac{9}{2}-9])$ $\displaystyle -([-\frac{9}{2}]-[-4])+([\frac{25}{2}-\frac{30}{2}]-[-\frac{9}{2}])$ $\displaystyle -(-\frac{9}{2}+4)+(-\frac{5}{2}+\frac{9}{2})$ $\displaystyle -(-\frac{1}{2})+(\frac{4}{2})$ $\displaystyle \frac{1}{2}+2$ $\displaystyle \frac{5}{2}$