Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 514: 17


$$\int_0^1{[-x+2 - (x^2-1)]dx}=\frac{13}{6}$$

Work Step by Step

Find region bounded by $y = x^2 -1, y= -x+2, x=0, x=1$ The formula we will use here is the following: $$\int_a^b {(top - bottom)}dx$$ We must identify which functions is on top versus on bottom for $x$ in $[0 , 1]$. By graphing both function on between $0$ and $1$. If we do so, we observer $y = -x+2$ is above $y = x^2 -1$, Hence: $$\int_0^1{[-x+2 - (x^2-1)]dx}$$ Simplify $$\int_0^1{(-x^2-x+3)dx}$$ We use a power rule to integrate the quadratic polynomial $$\frac{-x^3}{3} -\frac{x^2}{2}+3x |_0^1$$ Then evaluate the result $$\left(\frac{-(1)^3}{3} -\frac{(1)^2}{2}+3(1)\right) - \left( \frac{-(0)^3}{3} -\frac{(0)^2}{2}+3(0) \right) = \frac{13}{6}-0=\frac{13}{6}$$
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