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Question

Mathematics Question on Area under Simple Curves

The area of the region
\begin{array}{l} \left\\{\left(x,y\right);\left|x-1\right|\leq y \leq \sqrt{5-x^2} \right\\}\end{array}
is equal to

A

52sin−1⁡(35)−12\frac{5}{2}sin^{−1}⁡(\frac{3}{5})−\frac{1}{2}

B

5π4−32\frac{5π}{4}−\frac{3}{2}

C

3π4+32\frac{3π}{4}+\frac{3}{2}

D

5π4−12\frac{5π}{4}−\frac{1}{2}

Answer

5π4−12\frac{5π}{4}−\frac{1}{2}

Explanation

Solution

Area of the Region

A=∫−11(5−x2−(1−x))dx+∫12(5−x2−(x−1))dx\begin{array}{l} A=\displaystyle\int\limits_{-1}^1\left(\sqrt{5-x^2}-\left(1-x\right)\right)dx+\displaystyle\int\limits_{1}^2\left(\sqrt{5-x^2}-\left(x-1\right)\right)dx\end{array}
A=2(x25−x2+52sin⁡−1x5)−2x∣01+x25−x2+52sin⁡−1x5−x22+x∣12\begin{array}{l} \left.A=2\left(\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right)-2x\right|_0^1 \left.+\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}-\frac{x^2}{2}+x\right|_1^2\end{array}
=(5π4−12) sq. units\begin{array}{l} =\left(\frac{5\pi}{4}-\frac{1}{2}\right)\text{ sq. units} \end{array}