Question

The area bounded by the curves, $y=\sqrt{5-x^2}$ and $y=|x-1|$ is

• A. $\frac{5\pi -1}{4}$
• B. $\frac{5\pi +1}{4}$
• C. $\frac{5\pi -2}{4}$
• D. $\frac{5\pi -3}{4}$

Answer 1

The correct option is C $\frac{5\pi -2}{4}$

Given curve is

$y=\sqrt{5-x^2}$
$squaringbothsides$
$\Rightarrow y^2+x^2={\left(\sqrt{5}\right)}^2.......\left(i\right)$
$y=|x-1|$
$\Rightarrow y=x-1,y=1-x........\left(ii\right)$
$po\mathrm{int}of\mathrm{int}er\sec tionof\left(i\right)and\left(ii\right)$
${(x-1)}^2+x^2=5$
$\Rightarrow x^2+1-2x+x^2-5=0$
$\Rightarrow 2x^2-2x-4=0$
$\Rightarrow x^2-x-2=0$
$\Rightarrow (x-2)(x+1)=0$
$x=2,-1$
$\mathrm{Re}quiredarea$
$=\underset{-1}{\overset{2}{\int }}\sqrt{5-x^2}dx-\underset{-1}{\overset{2}{\int }}|x-1|dx$
$=\underset{-1}{\overset{2}{\int }}\sqrt{5-x^2}dx-\underset{-1}{\overset{1}{\int }}-(x-1)dx-\underset{1}{\overset{2}{\int }}(x-1)dx$
$={[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_{-1}^2-{[-\frac{x^2}{2}+x]}_{-1}^1-{[\frac{x^2}{2}-x]}_1^2$
$=[(\frac{2}{2}\sqrt{5-4}+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right))-(\frac{-1}{2}\sqrt{5-1}+\frac{5}{2}{\sin }^{-1}\left(\frac{-1}{\sqrt{5}}\right))]-[(-\frac{1}{2}+1)-(-\frac{1}{2}-1)]$
$-[(\frac{4}{2}-2)-(\frac{1}{2}-1)]$
$=[1+\frac{5}{2}{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+1+\frac{5}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)]-[(-\frac{1}{2}+1)-(-\frac{1}{2}-1)]-[(2-2)-(\frac{1}{2}-1)]$
$=[2+\frac{5}{2}[{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)]]-2-\frac{1}{2}$
$=\frac{5}{2}[{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)+{\cos }^{-1}\left(\frac{2}{\sqrt{5}}\right)]-\frac{1}{2}$
$=\frac{5}{2}\frac{\pi }{2}-\frac{1}{2}$
$=\frac{5\pi }{4}-\frac{1}{2}$
$=\frac{5\pi -2}{4}$