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Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 24 maths

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Answer:

y=\sin x, y=\sin 2 x \text { is in ratio } 2: 3

Hint:

Use two graph of \sin x.

Given:

Show that area under curve y=\sin x, y=\sin 2 x between x=0 and x=\frac{\pi }{3} are in ratio 2:3

Solution:

Area of graph 1 :

A_{1}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]

\begin{aligned} &A_{1}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{1}=\int_{0}^{\frac{\pi}{3}} \sin x d x \\\\ &A_{1}=[-\cos x]_{0}^{\frac{\pi}{3}} \end{aligned}

\begin{aligned} &A_{1}=\left[-\cos \frac{\pi}{3}+\cos 0\right] \\\\ &A_{1}=-\frac{1}{2}+1 \\\\ &A_{1}=\frac{1}{2} \end{aligned}

Area of graph 2 :                                        ............(i)

A_{2}=\int_{0}^{\frac{\pi}{3}}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]

\begin{aligned} &A_{2}=\int_{0}^{\frac{\pi}{3}} y d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} \sin 2 x d x \\\\ &A_{2}=\int_{0}^{\frac{\pi}{3}} 2 \sin x \cos x d x \end{aligned}

\begin{aligned} &A_{2}=2\left[\frac{-1}{4} \cos 2 x\right]_{0}^{\frac{\pi}{3}} \\\\ &A_{2}=\frac{1}{2}\left[-\cos 2 \frac{\pi}{3}+\cos 0\right] \end{aligned}

\begin{aligned} &A_{2}=\frac{1}{2}\left[1+\frac{1}{2}\right] \\\\ &A_{2}=\frac{1}{2}\left[\frac{3}{2}\right] \\\\ &A_{2}=\frac{3}{4} \end{aligned}                                            ............(ii)

From (i) and (ii)

\frac{A_{1}}{A_{2}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}

Thus area of curve y=\sin x, y=\sin 2 x for x=0 and x=\frac{\pi }{3} are in ratio 2:3

 

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