Question
$\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x$ बराबर है :
We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$
$\text { Hence } \int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} dx =\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$
$=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]$
$=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]$
$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$
$=24 \times \frac{\pi}{12}=2 \pi$
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$A_1=\left\{(x, y): x \geq 0, y \geq 0,2 x+2 y-x^2-y^2>1>x+y\right\}$
$A_2=\left\{(x, y): x \geq 0, y \geq 0, x+y>1>x^2+y^2\right\}$
$A_3=\left\{(x, y): x \geq 0, y \geq 0, x+y>1>x^3+y^3\right\}$
$A_1, A_2$, एवं $A_3$ के क्षेत्रफल क्रमशः $\left|A_1\right|,\left|A_2\right|$, एवं $\left|A_3\right|$ है, तब
तथा $\mathrm{f}(0)=0$ है। तो $\mathrm{f}\left(\frac{\pi}{2}\right)$ बराबर है