# How do you use integrals to find the area bounded by the curve y = (x^2 - 9) and the x-axis for x = 0 to x = 5?

Jul 27, 2015

I found: area$= - \frac{10}{3}$

#### Explanation:

You write it as:
${\int}_{0}^{5} \left({x}^{2} - 9\right) \mathrm{dx} =$
you can break it into two as:
${\int}_{0}^{5} \left({x}^{2}\right) \mathrm{dx} - {\int}_{0}^{5} \left(9\right) \mathrm{dx} =$
integrating you get:
${x}^{3} / 3 - 9 x {|}_{0}^{5}$

now you use you extremes of integration substituting them and subtracting the resulting expressions as:
$\left({5}^{3} / 3 - 9 \cdot 5\right) - \left({0}^{3} / 3 - 9 \cdot 0\right) =$
$= \frac{125}{3} - 45 = - \frac{10}{3}$

You may wonder about the negative sign but looking at your area:

you can see that quite a big chunk (pink) is below the $x$ axis giving a negative contribution (this are is equal to $- 18$) while the blue part (positive) gives you a contribution of only $14.6$.