Area Question 50
Question 50
- Sketch the region bounded by the curves $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ and find its area.
(1985, 5M)
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Solution:
- Given curves $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ could be sketched as shown, whose point of intersection are $5-x^{2}=(x-1)^{2}$
$\Rightarrow \quad 5-x^{2}=x^{2}-2 x+1$
$\Rightarrow \quad 2 x^{2}-2 x-4=0$ $\Rightarrow \quad x=2,-1$
$\therefore$ Required area
$=\int_{-1}^{2} \sqrt{5-x^{2}} d x-\int_{-1}^{1}(-x+1) d x-\int_{1}^{2}(x-1) d x$
$=\frac{x}{2} \sqrt{5-x^{2}}+\frac{5}{2} \sin ^{-1} \frac{x^{2}}{\sqrt{5}}-\frac{-x^{2}}{2}+x_{-1}^{1}-\frac{x^{2}}{2}-x_{1}^{2}$
$=1+\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}}–1+\frac{5}{2} \sin ^{-1} \frac{-1}{\sqrt{5}}$
$$ –\frac{1}{2}+1+\frac{1}{2}+1-2-2-\frac{1}{2}+1 $$
$=\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}}+\sin ^{-1} \frac{1}{\sqrt{5}}-\frac{1}{2}$
$=\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}} \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}} \sqrt{1-\frac{4}{5}}-\frac{1}{2}$
$=\frac{5}{2} \sin ^{-1}(1)-\frac{1}{2}=\frac{5 \pi}{4}-\frac{1}{2}$ sq units
