$\frac{{\displaystyle \underset{0}{\overset{}{}}}a(x)b(x)dx}{{\displaystyle \underset{0}{\overset{}{}}}b(x)dx}$

Hint: You should get $0\mathrm{.}34$ as your final answer. If you are unsure of where to start or get stuck, consider the following:

An integral can be calculated as the area under a curve. This may help you get the denominator of the desired expression $($hint: the denominator should be $1185).$

To tackle the numerator, consider applying the additive properties of definite integrals:

${\displaystyle \underset{a}{\overset{c}{}}}f(x)dx={\displaystyle \underset{a}{\overset{b}{}}}f(x)dx+{\displaystyle \underset{b}{\overset{c}{}}}f(x)dx$

This problem is easiest to solve if you break up the numerator on the intervals of $,$ and $9\mathrm{.}7.$ Then, remember the integral constant multiple rule, where if k is constant with respect to $x$ on the given interval between $g$ and $h,$ you can factor it out of an integral as: ${\displaystyle \underset{g}{\overset{h}{}}}kf(x)dx=k{\displaystyle \underset{g}{\overset{h}{}}}f(x)dx$

Using these two integral properties $($combined with the area$-$under$-$a$-$curve principle$),$ you can solve for the numerator. $($hint: the numerator should be $402).$