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Evaluate the following problems using second fundamental theorem: $\int\limits_{0}^{1}\sqrt{9-4x^{2}}dx$

$\begin{array}{1 1} \frac{\sqrt 5}{2}-\frac{9}{4} \sin ^{-1} \frac{2}{3} \\\frac{\sqrt 5}{2}+\frac{9}{4} \sin ^{-1} \frac{2}{3} \\ \frac{\sqrt 5}{2}+\frac{9}{4} \\\frac{\sqrt 7}{2}+\frac{9}{4} \sin ^{-1} \frac{2}{3} \end{array} $

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Toolbox:
  • $\int \sqrt {a^2-x^2}dx=\large\frac{a^2}{2} $$\sin ^{-1} \large\frac{x}{a}+\frac{x}{2} $$\sqrt {a^2-x^2}$
Given $\int \limits _0^1 \sqrt {9-4x^2}$
Step 1:
$\int \limits _0^1 \sqrt {9-4x^2}=\int \limits _0^1 \sqrt {3^2-(2x)^2}dx$
$\qquad=\int \limits _0^1 \sqrt {\bigg(\large\frac{3}{2}\bigg)^2 \normalsize -x^2}dx$
Step 2:
$\qquad=2 \bigg[\large\frac{x}{2} \sqrt {\bigg(\frac{3}{2}\bigg)^2-x^2}+ \large\frac{\bigg(\Large\frac{3}{2}\bigg)^2}{2}$$\sin ^{-1} \large\frac{x}{\Large\frac{3}{2}}\bigg]_0^1$
$\qquad=2 \bigg[\large\frac{x}{2} \sqrt {\bigg(\frac{3}{2}\bigg)^2-x^2}+ \large\frac{9}{8}$$\sin ^{-1} \large\frac{2x}{3}\bigg]_0^1$
$\qquad=2 \bigg[\large\frac{1}{2} \sqrt {\frac{9}{4} - \normalsize 1}+ \large\frac{9}{8}$$\sin ^{-1} \large\frac{2}{3}\bigg]$
$\qquad=\large\frac{\sqrt 5}{2}+\frac{9}{4}$$ \sin ^{-1} \large\frac{2}{3}$
answered Aug 14, 2013 by meena.p
 

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