University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 79

Answer

$\dfrac{1}{2}$

Work Step by Step

When the function $f(x)$ has minimum and maximum on $[a,b]$ then , we have $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ Given: $f(x)=\sin x \leq x$ for $x \geq 0$ or, $\sin x-x \leq 0$ for $x \geq 0$ Now, we have $\int_0^1 (\sin x-x) dx \leq 0$ or, $\int_0^1 \sin x dx -\int_0^1 x dx \leq 0$ or, $\int_0^1 \sin x dx \leq \int_0^1 x dx$ Thus, $\int_0^1 \sin x dx \leq \dfrac{1}{2}$ Hence, the upper bound is: $\dfrac{1}{2}$

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