Find the value of m so that the region above the line y=mx and below the parabola y=2x-x^2 has an area of 36 square units.

Answer 1

The first step is to find the points of intersection of the line and the parabola. The two solutions are:

x = 0 and x = 2-m.

The parabola is the overlying curve so

36 = ∫ dx [ 2 x - x² - mx ] with limits of integration: 0 and 2 - m.

The integral can be done term by term. There is no contribution from the lower limit so

36 = (2 - m)² - (2 -m)³ /3 - m (2-m)² /2

With some algebra, this can be rearranged as:

36 = (2-m)³ /6

so (2-m)³ = 6³ and m = -4