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

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

Answers (1)

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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