SATHEE: Area Question 50

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 $5 - x^{2} = (x - 1)^{2}$

$\Rightarrow 5 - x^{2} = x^{2} - 2 x + 1$

$\Rightarrow 2 x^{2} - 2 x - 4 = 0$ $\Rightarrow x = 2, - 1$

$∴$ Required area

$= \int_{- 1}^{2} \sqrt{5 - x^{2}} d x - \int_{- 1}^{1} (- x + 1) d x - \int_{1}^{2} (x - 1) d x$

$= \frac{x}{2} \sqrt{5 - x^{2}} + \frac{5}{2} \sin^{- 1} \frac{x^{2}}{\sqrt{5}} - \frac{- x^{2}}{2} + x_{- 1}^{1} - \frac{x^{2}}{2} - x_{1}^{2}$

$= 1 + \frac{5}{2} \sin^{- 1} \frac{2}{\sqrt{5}} - 1 + \frac{5}{2} \sin^{- 1} \frac{- 1}{\sqrt{5}}$

$- \frac{1}{2} + 1 + \frac{1}{2} + 1 - 2 - 2 - \frac{1}{2} + 1$

$= \frac{5}{2} \sin^{- 1} \frac{2}{\sqrt{5}} + \sin^{- 1} \frac{1}{\sqrt{5}} - \frac{1}{2}$

$= \frac{5}{2} \sin^{- 1} \frac{2}{\sqrt{5}} \sqrt{1 - \frac{1}{5}} + \frac{1}{\sqrt{5}} \sqrt{1 - \frac{4}{5}} - \frac{1}{2}$

$= \frac{5}{2} \sin^{- 1} (1) - \frac{1}{2} = \frac{5 π}{4} - \frac{1}{2}$ sq units

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