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  • Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 3

Need solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 3

Answers (1)

Answer:

\frac{8}{3}a^{2}\; sq\cdot units

Hint:

Find the shaded area

Given:

Find area of region bounded by parabola y^{2}=4 a x and the line x=a

Solution:

We have

                x=a                      … (i)                                                                                                                                                 

                y^{2}=4 a x            … (ii)                                                                                                                                      

Required area = shaded region OBAO

        \left[\because y^{2}=4 a x \Rightarrow y=\sqrt{4 a x}\right]

=2(\text { Shaded region } \mathrm{OBCO}) \quad\left[\because \int_{a}^{b} x^{n} d x=\left[\frac{x^{n+1}}{n+1}\right]_{a}^{b}\right]

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

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

    \begin{aligned} &=\frac{4 \sqrt{4 a}}{3}\left[(a)^{\frac{3}{2}}-(0)^{\frac{3}{2}}\right] \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a})^{3} \\\\ &=\frac{4 \sqrt{4 a}}{3} \times(\sqrt{a} \times \sqrt{a} \times \sqrt{a}) \end{aligned}

    \begin{aligned} &=\frac{4 \sqrt{4 a}}{3} \times a \sqrt{a} \\\\ &=\frac{8 a^{2}}{3} \text { sq.unit } \end{aligned}

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