University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 6


$\int_{0}^{1} \sqrt{4-x^2}dx$.

Work Step by Step

By using the definition of the definite integral, P is a partition of [0,1], therefore the lower and upper limits of the integration are 0 and 1. $f(c_{k})=\sqrt{4-c_{k}^2}$ is the function in the additive of the Riemann sums, therefore $f(x)=\sqrt{4-x^2}$. Therefore the solution is: $\int_{0}^{1} \sqrt{4-x^2}dx$.
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