Question

Find the area bounded by the curve $y=9-x^2$ and the $X$ -axis.

Answer 1

$\to y=-x^2+9$ , Graphing $y=-x^2+9$

$\to A={\int }_{x_1}^{x_2}(-x^2+9)dx={\int }_{-3}^3(-x^2+9)dx$ It is a concave down parabola

$A={[\frac{-x^3}{3}+9x]}_{-3}^3$

$A=[{\frac{-3}{3}}^3+9(3)-\int -\frac{(-3{)}^3}{3}+9(-3)]$

$A=|{\frac{-2.3}{3}}^3+2(9\left)\right(3\left)\right|=|-18+18\left(3\right)|=36$