Q116 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

$$ \int_0^2 \frac{x}{(2-x)^{\frac{3}{4}}} d x= $$

  • A
    $\frac{24}{5} 2^{\frac{1}{4}}$
  • B
    $\frac{5}{24} 2^{\frac{3}{4}}$
  • Correct answer
    C
    $\frac{32}{5} 2^{\frac{1}{4}}$
  • D
    $\frac{5}{12} 2^{\frac{3}{4}}$

Solution

$\int_0^2 \frac{x}{(2-x)^{3 / 4}} d x$ Let $2-x=t$ $$ \begin{aligned} & =-\int_2^0 \frac{(2-t) d t}{(t)^{3 / 4}}=-\int_2^0 \frac{2}{t^{3 / 4}} d t+\int_2^0 \frac{t}{t^{3 / 4}} d t \\ & =\int_2^0 2 t^{-3 / 4} d t+\int_2^0 t^{1 / 4} d t \\ & =-2 \cdot 4\left[t^{1 / 4}\right]_2^0+\frac{4}{5}\left[t^{5 / 4}\right]_2^0 \\ & =-8(0-\sqrt[4]{2})+\frac{4}{5}\left(0-(2)^{5 / 4}\right) \\ & =8 \cdot \sqrt[4]{2}+\left[-\frac{4}{5}(2)^{5 / 4}\right]=\sqrt[4]{2}\left(8-\frac{8}{5}\right)=\frac{32}{5} \cdot \sqrt[4]{2} . \end{aligned} $$

Hence, the correct answer is option: 'C'.

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