Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.2 Definite Integrals - 5.2 Exercises: 9


The formula is $$\int_0^axdx=\frac{a^2}{2}.$$

Work Step by Step

Note that the integrand $x$ is nonnegative on the interval of integration $[0,a]$ so this integral will gives simply the area of the the triangular region bounded by the graph of the function, the $x$ axis and a vertical line $x=a$ (shown on the figure below). The area of this right triangle is just the product of its legs times one half. Since $f(x)=x$ then $f(a)=a$ so the vertical leg has the same length as the horizontal leg which is $a$ so we have $$\int_0^af(x)dx=\frac{1}{2}a\cdot a=\frac{a^2}{2}.$$
Small 1511131916
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.