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  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 35 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 35 maths textbook solution.

Answers (1)

Answer: \frac{720}{7}

Hint: Use indefinite integral formula and the given limits to solve this integral.

Given: \int_{2}^{12}x(x-4)^{\frac{1}{4}}dx

Solution: \int_{2}^{12}x(x-4)^{\frac{1}{4}}dx

Put   x - 4 = t^3 \Rightarrow dx = 3t^2dt

When x = 4 then t =0 & when x =12 then t = 2

\begin{aligned} &\therefore \int_{4}^{12} x(x-4)^{1 / 3} d x \\ &=\int_{0}^{2}\left(t^{3}+4\right)\left(t^{3}\right)^{1 / 3} \cdot 3 t^{2} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because x-4=t^{3} \Rightarrow t^{3}+4=x\right] \\ &=\int_{0}^{2}\left(t^{3}+4\right) t \cdot 3 t^{2} d t \\ &=\int_{0}^{2}\left(t^{3}+4\right) 3 t^{3} d t \end{aligned}

\begin{aligned} &=3 \int_{0}^{2}\left(t^{3+3}+4 t^{3}\right) d t=3 \int_{0}^{2}\left(t^{6}+4 t^{3}\right) d t\\ &=3 \int_{0}^{2} t^{6} d t+12 \int_{0}^{2} t^{3} d t\\ &=3\left[\frac{t^{6+1}}{6+1}\right]_{0}^{2}+12\left[\frac{t^{3+1}}{3+1}\right]_{0}^{2}\ \! \! \! \! \! \! &\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]\\ &=\frac{3}{7}\left[t^{7}\right]_{0}^{2}+\frac{12}{4}\left[t^{4}\right]_{0}^{2} \end{aligned}

\begin{aligned} &=\frac{3}{7}\left[2^{7}-0^{7}\right]+3\left[2^{4}-0^{4}\right] \\ &=\frac{3}{7}[128-0]+3[16-0] \\ &=3\left[\frac{128}{7}+16\right] \\ &=3\left[\frac{128+112}{7}\right] \\ &=3 \times \frac{240}{7} \\ &=\frac{720}{7} \end{aligned}

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