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

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}