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  • Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 28

Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 28

Answers (1)

Answer:

\frac{1}{2756}

Hint:

Integrate then use limits

Given:

\int_{0}^{1} \frac{x}{(1-x)^{54}} d x

Solution:

\begin{aligned} &I=\int_{0}^{1} \frac{x}{(1-x)^{54}} d x \\\\ &I=\int_{0}^{1} \frac{x}{(x-1)^{54}} \quad\left[\mathrm{Q}(x-y)^{2}=(y-x)^{2}\right] \end{aligned}

    \begin{aligned} &=\int_{0}^{1} \frac{x+1-1}{(x-1)^{54}} d x \\\\ &=\int_{0}^{1} \frac{(x-1)}{(x-1)^{54}} d x+\int_{0}^{1} \frac{1}{(x-1)^{54}} d x \end{aligned}

    \begin{aligned} &=\int_{0}^{1}(x-1)^{-53} d x+\int_{0}^{1}(x-1)^{-54} d x \\\\ &=\left[-\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[-\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \end{aligned}

    \begin{aligned} &=\left[\frac{(x-1)^{-52}}{52}\right]_{0}^{1}+\left[\frac{(x-1)^{-53}}{53}\right]_{0}^{1} \\\\ &=\frac{(-1)^{52}}{52}+\frac{(-1)^{53}}{53} \end{aligned}

    \begin{aligned} &=\frac{1}{52}-\frac{1}{53}=\frac{1}{52 \times 53} \\\\ &=\frac{1}{2756} \end{aligned}

 

 

Posted by

infoexpert26

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