Question

12. Find the area lying above the x - axis and below the parabolic curve y = 4x - x2.

Answer 1

First we find the zeroes so we don't take the integral of negative bits

4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4

$\int\limits^4_0 {4x-x^2} \, dx =[2x^2- \frac{1}{3}x^3]^4_0=(32- \frac{64}{3})-(0)= 10.6666666666$ or aout 10 and 2/3

C is answer