Find the area enclosed between C, the curve y=6x-x^2, L, the line y=16-2x and the y axis.

First we need to find the intersection point(s) of L and C so set 6x-x²=16-2x and rearrange to get x²-8x+16=0 so (x-4)²=0.Repeated root so line is tangent to the curve at x=4, y=16-2(4)=8 that is the point (4,8).Area= integral between 0 and 4 (16-2x) dx - integral between 0 and 4 (6x-x²) dx= [16x - x² - 3x² + x³/3] evaluated between 0 and 4= 64/3