Mathematics

# Solve $\int\limits_2^6 {\sqrt {\left( {6 - x} \right)\left( {x - 2} \right)} dx}$

##### SOLUTION

We have,

$I=\int_{2}^{6}{\sqrt{\left( 6-x \right)\left( x-2 \right)}}\,dx$

$I=\int_{2}^{6}{\sqrt{6x-12-{{x}^{2}}+2x}}\,dx$

$I=\int_{2}^{6}{\sqrt{8x-12-{{x}^{2}}}}\,dx$

$I=\int_{2}^{6}{\sqrt{8x-12-16+16-{{x}^{2}}}}\,dx$

$I=\int_{2}^{6}{\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( 4-x \right)}^{2}}}}\,dx$

Let

$t=4-x$

$-dt=dx$

Therefore,

$I=-\int_{2}^{-2}{\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{t}^{2}}}}\,dt$

We know that

$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\dfrac{1}{2}x\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$

Therefore,

$I=-\left[ \dfrac{1}{2}t\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{t}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{t}{2\sqrt{7}} \right) \right]_{2}^{-2}$

$=-\left( \dfrac{1}{2}\left( -2 \right)\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( -2 \right)}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{-2}{2\sqrt{7}} \right) \right)+$

$\left( \dfrac{1}{2}\left( 2 \right)\sqrt{{{\left( 2\sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}}+\dfrac{1}{2}{{\left( 2\sqrt{7} \right)}^{2}}{{\sin }^{-1}}\left( \dfrac{2}{2\sqrt{7}} \right) \right)$

$I=- \left( -\sqrt{28-4}+\dfrac{1}{2}\times 28{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right) \right)+\left( \sqrt{28-4}+\dfrac{1}{2}\times 28{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right) \right)$

$I=-\left( -\sqrt{24}+14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right) \right)+\left( \sqrt{24}+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right) \right)$

$I=2\sqrt{24}-14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right)+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right)$

Hence, the value is $2\sqrt{24}-14{{\sin }^{-1}}\left( \dfrac{-1}{\sqrt{7}} \right)+14{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{7}} \right)$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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