Find the area of the shaded region.

Question

The x y coordinate plane is given. There is a curve labeled y = x³, a line labeled y = x + 6, a line labeled y = −1⁄2 x, and a region on the graph. 

1. The curve enters the window in the third quadrant, goes up and right becoming less steep, becomes nearly horizontal at the origin (where it crosses the second line), goes up and right becoming more steep, passes through the point (2, 8) crossing the first line, and exits the window in the first quadrant. 
2. The first line enters the window in the third quadrant, goes up and right, crosses the x-axis at x = −6, passes through the point (−4, 2) crossing the second line, crosses the y-axis at y = 6,passes through the point (2, 8) crossing the curve, exits at the point (4, 10), and exits the window in the first quadrant. 
3. The second line enters the window in the second quadrant, goes down and right, passes through the point (−4, 2) crossing the first line, crosses the x-axis at the origin, and exits the window in the fourth quadrant. 
4. The region is the area enclosed by the curve and the lines.

Answer 1

Divide the shaded region into two parts by creating the segment from (0,0) to (0,6). Call the region to the left of that segment R₁ and to the right of that segment R₂.

The area of R₂ can be found using calculus, but also by the fact that that region is triangle with a base equal to 6 and a height equal to 4 (picture the segment from (-4,2) to (0,2) as its altitude. The area of that region is

A₁ = (1/2)(6)(4) = 12.

You could also use calculus:

A₁= ∫_-4⁰ [(x+6) - (-1x/2)]dx = 3/4 x² + 6x |_-4⁰ = -[3/4 (16) + 6(-4)] = -[-12] = 12.

For R₂ you will have to use calculus. For each x between 0 and 2 picture a thin rectangle with y-coordinate at the top (x+6) and y-coordinate at the bottom (x³). The height of that rectangle can be represented as (x+6) - x³.

A₂ = ∫₀²[x + 6 - x³]dx = (1/2)x² + 6x -(1/4)x⁴ |₀² = [(2 + 12 - 4) - (0)] = 10

The total area of the region is: 12 + 10 = 22 square units.