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  • Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 8 maths

Explain solution RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 8 maths

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best_answer

Answer:

\frac{56}{9}sq\cdot units

Hint:

y=\sqrt{6 x+4} \quad \text { at } x=0, x=2

Given:

Find area under curve y=\sqrt{6 x+4} above x-axis form x=0 to x=2 . Draw sketch of curve also.

Solution:

y=\sqrt{6 x+4}  represent a parabola with vertex v=\left(\frac{-2}{3}, 0\right) and symmetric aboutx-axis where x=0 is y-axis

The rectangle move from x=0 to x=2

Consider,                

        \begin{aligned} &\text { Area } O A B C=\int_{0}^{2}|y| d x \\\\ &A=\int_{0}^{2} \sqrt{6 x+4} d x \\\\ &A=\int_{0}^{2}(6 x+4)^{\frac{1}{2}} d x \end{aligned}                                                            

        \begin{aligned} &A=\frac{1}{6}\left[\frac{(6 x+4)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\\\ &A=\frac{2}{18}\left[(16)^{\frac{3}{2}}-(4)^{\frac{3}{2}}\right] \end{aligned}

        \begin{aligned} &A=\frac{2}{18}\left[4^{3}-2^{3}\right] \\\\ &A=\frac{2}{18}[64-8] \\\\ &A=\frac{2}{18}[56] \\\\ &A=\frac{56}{9} s q \cdot \text { unit } \end{aligned}

       

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