Area under a parabola and mean ordinate

Text

Compute $\displaystyle\int_{0}^{5}\left(x^{2}+2x-4\right)dx$ and the mean ordinate of the function over $[0,5]$.

Solution

$\int_{0}^{5}\!\left(x^{2}+2x-4\right)dx=\left[\frac{x^{3}}{3}+x^{2}-4x\right]_{0}^{5}=\frac{125}{3}+5=\frac{140}{3}\approx46.667.$ The mean ordinate is $\displaystyle\bar y=\frac{1}{5}\int_{0}^{5}(\dots)\,dx=\frac{140/3}{5}=\frac{28}{3}\approx9.333.$ $\boxed{A=\dfrac{140}{3}\approx46.67,\qquad \bar y=\dfrac{28}{3}\approx9.33}$