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  • Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 12 maths

Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 12 maths

Answers (1)

Answer:

\frac{\pi}{2} \; sq\cdot unit

Hint:

\left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]

Given:

Draw a rough sketch of graph of function y=2 \sqrt{1-x^{2}}, x \in[0,1] and evaluate the area enclosed between the curve and x-axis

Solution:

We have

\begin{aligned} &y=2 \sqrt{1-x^{2}} \\\\ &\frac{y}{2}=\sqrt{1-x^{2}} \\\\ &\frac{y^{2}}{4}=1-x^{2} \end{aligned}

\begin{aligned} x^{2}+\frac{y^{2}}{4} &=1 \\\\ \frac{x^{2}}{1}+\frac{y^{2}}{4} &=1 \end{aligned}

Since given equation \frac{x^{2}}{1}+\frac{y^{2}}{4}=1

\begin{aligned} &A=\int_{0}^{1} 2 \sqrt{1-x^{2}} d x \\\\ &A=2 \int_{0}^{1} \sqrt{1-x^{2}} d x \\\\ &A=2\left[\frac{1}{2} x \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{0}^{1} \end{aligned}

\begin{aligned} &A=2\left[\frac{1}{2} \sin ^{-1} 1\right] \\\\ &A=\sin ^{-1} 1 \\\\ &A=\frac{\pi}{2} s q \cdot \text { unit } \end{aligned}           \left[\because \int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c\right]

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