## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 64

#### Answer

$$\frac{2}{15}$$

#### Work Step by Step

Since $$\int_{1}^{b} f(x) d x =1-b^{-1}$$ Then \begin{aligned} \int_{1}^{5} f(x) d x&=\int_{1}^{3} f(x) d x+\int_{3}^{5} f(x) d x\\ \end{aligned} Solve for the last integral: \begin{aligned} & \int_{3}^{5} f(x) d x=\int_{1}^{5} f(x) d x-\int_{1}^{3} f(x) d x\\ &=\left(1-5^{-1}\right)-\left(1-3^{-1}\right)\\ &=\left(1-\frac{1}{5}\right)-\left(1-\frac{1}{3}\right)\\ &=\frac{4}{5}-\frac{2}{3}\\ &=\frac{2}{15} \end{aligned}

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