Question

Sketch the region enclosed by the given curves and

find its area. 20. \(y = \frac{1}{4}{x^2},y = 2{x^2},x + y = 3,x \ge 0\).

Short answer

The area is \(\frac{3}{2}\).

Step-by-step solution

The sketch of the graph of given equations

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

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