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Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 30

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Answer:

16^{\frac{1}{3}}

Hint:

Use integration.

Given:

If area between curve x=y^{2} and x=4 divide two equal part of line x=a .find using integration.

Solution:

Given curve       

y^{2}=x

Let AB represent line x=a

    CD  represent line x=4

Since line x=a divide the region in two equal parts

Area of OBA=  Area ofABCD

\begin{aligned} &2 \int_{0}^{a} y d x=2 \int_{a}^{4} y d x \\\\ &\int_{0}^{a} y d x=\int_{a}^{4} y d x \end{aligned}                        .............(i)

Now, y^{2}=x \Rightarrow y=\pm \sqrt{x}

y=\sqrt{x}

From (i)

\int_{0}^{a} y d x=\int_{a}^{4} y d x

\begin{aligned} &\int_{0}^{a} \sqrt{x} d x=\int_{a}^{4} \sqrt{x} d x \\\\ &\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{0}^{a}=\left(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right)_{a}^{4} \end{aligned}

\begin{aligned} &{\left[x^{\frac{3}{2}}\right]_{0}^{a}=\left[x^{\frac{3}{2}}\right]_{a}^{4}} \\\\ &a^{\frac{3}{2}}=4^{\frac{3}{2}}-a^{\frac{3}{2}} \\\\ &2 a^{\frac{3}{2}}=4^{\frac{3}{2}} \end{aligned}

Take \frac{2}{3}^{t h}root on both sides        

\begin{aligned} &2^{\frac{2}{3}} \cdot a^{\frac{3}{2} \cdot \frac{2}{3}}=4^{\frac{3}{2} \frac{2}{3}} \\\\ &2^{\frac{2}{3}} \cdot a=4 \\\\ &a=\frac{2^{2}}{2^{\frac{2}{3}}} \end{aligned}

\begin{aligned} &a=2^{2-\frac{2}{3}}=2^{\frac{6-2}{3}} \\\\ &a=2^{\frac{4}{3}} \\\\ &a=\left(2^{2}\right)^{\frac{2}{3}}=4^{\frac{2}{3}} \end{aligned}

\begin{aligned} &=\left(4^{2}\right)^{\frac{1}{3}}=16^{\frac{1}{3}} \\\\ &a=16^{\frac{1}{3}} \end{aligned}

 

 

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