Find the exact value of the volume of a solid whose base is bounded by the graphs of y=x+1 and y=x^2-1 with cross-sections that are rectangles of height 1, taken perpendicular to the x-axis.

Answer 1

The base of our solid would be (x+1) - (x²-1) = -x² + x + 2

Given our cross-sections are rectangles perpendicular to the x-axis with a height of 1, then the integral we'll need to compute is:

V = ∫[-1,2] (-x²+x+2) dx

V = -x³/3 + x²/2 + 2x [-1,2]

V = [-(2)³/3 + (2)²/2 + 2(2)] - [-(-1)³/3 + (-1)²/2 + 2(-1)]

V = [-8/3 + 2 + 4] - [1/3 + 1/2 - 2]

V = [-16/6 + 12/6 + 24/6] - [2/6 + 3/6 - 12/6]

V = 20/6 - (-7/6)

V = 27/6

V = 9/2

Thus, the exact value of the volume of the solid would be 9/2 cubic units.

Answer 2

x +1 = x² - 1; x² - x - 2 = 0; (x - 2)(x + 1) = 0; x = - 1 or x = 2; area A(x) = [(x + 1) - (x² - 1)]·1 = x + 2 - x²;

Volume v = ∫_-1²(x + 2 - x²)dx = (x²/2 + 2x - x³/3)_-1² = (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = 4.5