Answer to Question #176649 in Calculus for Joshua

Question

Find the area bounded by the curve y=4x-x^2 and the lines x=-2 and y=4.

Answer 1

Let us find the area bounded by the curve $y=4x-x^2$ and the lines $x=-2$ and $y=4$. Firstly, let us sketch the graph:

It follows that that the area $A$ is:

$A=\int_{-2}^2(4-(4x-x^2))dx=\int_{-2}^2(x^2-4x+4)dx=(\frac{x^3}{3}-2x^2+4x)|_{-2}^2= \frac{8}{3}-8+8-(-\frac{8}{3}-8-8)=\frac{8}{3}+\frac{8}{3}+16=21\frac{1}{3}$

Answer: $21\frac{1}{3}$ sq. units