Question

${\int }_0^1\frac{x}{(1-x{)}^{3/4}}dx$ is equal to:

• A. $\frac{-12}{5}$
• B. $\frac{16}{5}$
• C. $\frac{-16}{5}$
• D. none of these

Answer 1

Answer: (B)

If the product of two functions is given, we can integrate by using the method of integration by parts.
$<br/>\begin{array}{l}<br/>\text{ Let }I={\int }_0^{1}x(1-x{)}^{-3/4}dx\\ <br/>=\frac{x(1-x{)}^{1/4}(-1)}{1/4}+\int \frac{(1-x{)}^{1/4}}{1/4}dx\\ <br/>={\left[4x(1-x{)}^{1/4}-\frac{4(1-x{)}^{5/4}}{5/4}\right]}_0^1\\ <br/>=\left[0+0-\left(0-\frac{16}{5}\right)\right]\\ <br/>=\frac{16}{5}<br/>\end{array}<br/>$
Alternate Solution:
$<br/>\begin{array}{l}<br/>\text{ Let }I={\int }_0^{1}x(1-x{)}^{-3/4}dx\\ <br/>={\int }_0^1(1-x)[1-(1-x)]dx\\ <br/>={\int }_0^1(1-x)x^{-3/4}dx\\ <br/>={\int }_0^1\left(x^{-3/4}-x^{1/4}\right)dx\\ <br/>={\left[\frac{x^{1/4}}{1/4}-\frac{x^{5/4}}{5/4}\right]}_0^1\\ <br/>=\left[4-\frac{4}{5}\right]\\ <br/>=\frac{16}{5}<br/>\end{array}<br/>$