Search any question & find its solution
Question:
Answered & Verified by Expert
$$
\int_0^2 \frac{x}{(2-x)^{\frac{3}{4}}} d x=
$$
Options:
\int_0^2 \frac{x}{(2-x)^{\frac{3}{4}}} d x=
$$
Solution:
1405 Upvotes
Verified Answer
The correct answer is:
$\frac{32}{5} 2^{\frac{1}{4}}$
$\int_0^2 \frac{x}{(2-x)^{3 / 4}} d x$
Let $2-x=t$
$$
\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}
$$
Let $2-x=t$
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.