Question

Find the area of the closed figure bounded by the following curve
$y=4x-x^2$ and the $x$ axis.

Answer 1

Given the curve $y=4x-x^2$

$0=4x-x^2$

$x=0,4$

Area enclosed between x=0 and x=4 is ${\int }_0^4(4x-x^2-0)$

${\int }_0^4(4x-x^2)=[2x^2-\frac{x^3}{3}{]}_0^4=32-32/3=\frac{64}{3}$