Consider the region R that is between the graphs of y = x2-1, y = -x+1 on the interval x = 0 to x = 2. Find the area of this region

Answered: Consider the region R that is between…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter10: Analytic Geometry

Section10.1: The Rectangular Coordinate System

Problem 40E: Find the exact volume of the solid that results when the region bounded in quadrant I by the axes...

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Consider the region R that is between the graphs of y = x²-1, y = -x+1 on the interval x = 0 to x = 2.

Find the area of this region

Expert Solution

Step 1

The area between the graphs of $y = x^{2} - 1$ and $y = - x + 1$ on the interval $x = 0$ to $x = 2$ will be:

$\begin{array}{rcl} A & = & \int_{0}^{2} |(x^{2} - 1) - (- x + 1)| d x \\ = & \int_{0}^{2} |x^{2} + x - 2| d x \\ = & \int_{0}^{1} [- (x^{2} + x - 2)] d x + \int_{1}^{2} [x^{2} + x - 2] d x \\ = & {[- \frac{x^{3}}{3} - \frac{x^{2}}{2} + 2 x]}_{0}^{1} + {[\frac{x^{3}}{3} + \frac{x^{2}}{2} - 2 x]}_{1}^{2} \\ = & [(- \frac{1^{3}}{3} - \frac{1^{2}}{2} + 2 (1)) - (- \frac{0^{3}}{3} - \frac{0^{2}}{2} + 2 (0))] + [(\frac{2^{3}}{3} + \frac{2^{2}}{2} - 2 (2)) - (\frac{1^{3}}{3} + \frac{1^{2}}{2} - 2 (1))] \\ = & [(- \frac{1}{3} - \frac{1}{2} + 2) - 0] + [(\frac{8}{3} + 2 - 4) - (\frac{1}{3} + \frac{1}{2} - 2)] \\ = & 3 \end{array}$

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