Find the area of the closed figure bounded by the following curves
$y$ = $x^{2}, y = 1 + \frac{1}{4} x^{2}$ .

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Solution

Given the curves $y = x^{2}, y = 1 + \frac{1}{4} x^{2}$
$x^{2} = 1 + \frac{1}{4} x^{2} \Rightarrow x = \pm \frac{2}{\sqrt{3}}$
The area enclosed between $x = - \frac{2}{\sqrt{3}} a n d x = \frac{2}{\sqrt{3}}$
$\int_{- \frac{2}{\sqrt{3}}}^{\frac{2}{\sqrt{3}}} (- x^{2} + 1 + \frac{1}{4} x^{2}) d x = [- \frac{x^{3}}{4} + x]_{- 2 / \sqrt{3}}^{2 / \sqrt{3}} = - \frac{4}{3 \sqrt{3}} + \frac{4}{\sqrt{3}} = \frac{8}{3 \sqrt{3}} s q . u n i t s$

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