Answer to Question #151571 in Calculus for Joshua

Question

Find the area bound by the curves y=-x^2+4 and y=0.

Answer 1

Solve

$$-x^2+4=0$$

$$x_1=-2, x_2=2$$

We see that $y=-x^2+4\geq0$ for $-2\leq x\leq2.$ Then

$$Area=\displaystyle\int_{-2}^2(-x^2+4)dx=\big[-\dfrac{x^3}{3}+4x\big]\begin{matrix} 2 \\ -2 \end{matrix}$$

$$=-\dfrac{8}{3}+8-(\dfrac{8}{3}-8)=\dfrac{32}{3}(units^2)$$

$\dfrac{32}{3}$ square units