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Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 39 maths textbook solution.

Answers (1)

Answer    : \frac{4\sqrt{2}}{3}

Hint         :use indefinite integral formula and the limits to solve this integral

Given      : \int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x

Solution :-\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x
 put x^5+1=t \Rightarrow 5x^4dx=dt
when x= -1 then t=0 and when x=1 then t=2

\begin{aligned} &\int_{-1}^{1} 5 x^{4} \sqrt{x^{5}+1} d x \\ &=\int_{0}^{2} \sqrt{t} d t=\int_{0}^{2} t^{\frac{1}{2}} d t \\ &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2} \\ &=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2}=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{2}=\frac{2}{3}\left[2^{\frac{3}{2}}-0\right] \\ &=\frac{2}{3}(\sqrt{2})^{2 \times \frac{3}{2}}=\frac{2}{3}(\sqrt{2})^{3}=\frac{2 \times 2 \sqrt{2}}{3}=\frac{4 \sqrt{2}}{3} \end{aligned}


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