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Essential Calculus: Early Transcendentals
Found in: Page 289
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Calculate the area of the region that lies under the curve and above the x-axis.

\({\rm{y = 2x - }}{{\rm{x}}^{\rm{2}}}\)

Area of the region is \(\frac{{\rm{4}}}{{\rm{3}}}\)

See the step by step solution

Step by Step Solution

Step 1: To find the x-intercepts:

The area of region under a curve by the definite integral between the x-intercepts.

To find the x-intercepts:

\(\begin{aligned}{l}{\rm{2x - }}{{\rm{x}}^{\rm{2}}}{\rm{ &= 0}}\\{\rm{x(2 - x) &= 0}}\\{\rm{x &= 0,x &= 2}}\end{aligned}\)

Step 2: Evaluating the definite integral.

\(\begin{aligned}{c}\int_{\rm{0}}^{\rm{2}} {\left( {{\rm{2x - }}{{\rm{x}}^{\rm{2}}}} \right)} {\rm{dx}}\\{\rm{ &= }}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{\rm{2}}\\{\rm{ &= 4 - }}\frac{{\rm{8}}}{{\rm{3}}}{\rm{ &= }}\frac{{\rm{4}}}{{\rm{3}}}\end{aligned}\)

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