Answer to Question #344324 in Calculus for Jane

Question

3) Solve the area bounded by the curve y = 4x-x and the lines x = -2 and y = 4.

Answer 1

$$4x-x^2=4$$

$$(x-2)^2=0$$

$$x=2$$

$$A=\displaystyle\int_{-2}^{2}(4-(4x-x^2))dy$$

$$=\big[4x-2x^2+\dfrac{x^3}{3}\big]\begin{matrix} 2 \\ -2 \end{matrix}$$

$$=8-8+\dfrac{8}{3}+8+8+\dfrac{8}{3}=\dfrac{64}{3}({units}^2)$$