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  • Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 47.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 47.

Answers (1)

Answer :   \frac{19}{99}

Hint: Use in definite formula and the given limits to solve this integral

Given:

\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx

Solution:

\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx

Put (30-x^{\frac{3}{2}})=t

\begin{aligned} &\Rightarrow-\frac{3}{2} x^{\left(\frac{3}{2}-1\right)} d x=d t \\ &\Rightarrow-\frac{3}{2} x^{\frac{1}{2}} d x=d t \\ &\Rightarrow-\frac{2}{3 \sqrt{x}} d t=d x \end{aligned}

 

when x=4  then t=22

 when x=9  then t=3  then

\begin{aligned} &\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{3 / 2}\right)^{2}} d x \\ &=\int_{22}^{3} \frac{\sqrt{x}}{t^{2}} \cdot \frac{-2 d t}{3 \sqrt{x}} \\ &=\frac{-2}{3} \int_{22}^{3} t^{2} d t \\ &=-\frac{2}{3}\left[\frac{t^{-2+1}}{-2+1}\right]_{22}^{3} \end{aligned}

\begin{aligned} &=\frac{-2}{3}\left[\frac{t^{-1}}{-1}\right]_{22}^{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{3}\left[\frac{1}{t}\right]_{22}^{3} \\ &=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right] \\ &=\frac{2}{3}\left[\frac{22-3}{66}\right] \\ &=\frac{2}{3} \cdot \frac{19}{66} \\ &=\frac{19}{99} \end{aligned}

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