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Found in: Page 369

### Essential Calculus: Early Transcendentals

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

# Sketch the region enclosed by the given curves andfind its area. 20. $$y = \frac{1}{4}{x^2},y = 2{x^2},x + y = 3,x \ge 0$$.

The area is $$\frac{3}{2}$$.

See the step by step solution

## Simplify the given equations

$$y = \frac{1}{4}{x^2},y = 2{x^2},x + y = 3,x \ge 0$$

The parabolas intersect at $$(0,0)$$. Find where the line$$(y = 3 - x)$$intersects the parabolas.

\begin{aligned}{l}\frac{1}{4}{x^2} = 3 - x\\\frac{1}{4}{x^2} + x - 3 = 0\\{x^2} + 4x - 12 = 0\\(x + 6)(x - 2) = 0\\x = - 6,2\end{aligned}

\begin{aligned}{l}2{x^2} = 3 - x\\2{x^2} + x - 3 = 0\\(2x + 3)(x - 1) = 0\\x = - \frac{3}{2},1\end{aligned}

Because we are given that $$x \ge 0\;so\,x = 1,2$$

## Area calculation

The integral must be done in two sections because the top function changes at x=1. From x=0 to 1, the top on is $$y = 2{x^2}$$,from x=1 to 2 the top is $$y = 3 - x$$. The bottom one is the same for both sections.

\begin{aligned}{l}A = \int\limits_0^1 {\left( {2{x^2} - \frac{1}{4}{x^2}} \right)dx + } \int\limits_1^2 {\left( {3 - x - \frac{1}{4}{x^2}} \right)dx} \\A = \int\limits_0^1 {\left( {\frac{7}{4}{x^2}} \right)dx + } \int\limits_1^2 {\left( {3 - x - \frac{1}{4}{x^2}} \right)dx} \\A = \left( {\frac{7}{4}\left( {\frac{1}{3}{x^3}} \right)} \right)_0^1 + \left( {3x - \frac{1}{2}{x^2} - \frac{1}{4}\left( {\frac{1}{3}{x^3}} \right)} \right)_1^2\\A = \left( {\frac{7}{{21}} - 0} \right) + \left( {3(2) - \frac{1}{2}{{(2)}^2} - \frac{1}{{12}}{{(2)}^3} - \left( {3 - \frac{1}{2} - \frac{1}{{12}}} \right)} \right)\end{aligned}

## Further simplification

\begin{aligned}{l} = \frac{7}{{12}} + \left( {6 - 2 - \frac{8}{{12}} - 3 + \frac{1}{2} + \frac{1}{{12}}} \right)\\ = \frac{7}{{12}} + 1 - \frac{8}{{12}} + \frac{1}{2} + \frac{1}{{12}}\\ = \frac{{7 + 12 - 8 + 6 + 1}}{{12}} = \frac{{18}}{{12}}\\A = \frac{3}{2}\end{aligned}