A) \[\frac{-12}{5}\]
B) \[\frac{16}{5}\]
C) \[\frac{-16}{5}\]
D) none of these
Correct Answer: B
Solution :
Key Idea: If the product of two functions is given, we can integrate by using the method of integration by parts. Let \[I=\int_{0}^{1}{x{{(1-x)}^{-3/4}}}dx\] \[=\frac{x{{(1-x)}^{1/4}}(-1)}{1/4}+\int_{{}}^{{}}{\frac{{{(1-x)}^{1/4}}}{1/4}}dx\] \[=\left[ 4x{{(1-x)}^{1/4}}-\frac{4{{(1-x)}^{5/4}}}{5/4} \right]_{0}^{1}\] \[=\left[ 0+0-\left( 0-\frac{16}{5} \right) \right]\] \[=\frac{16}{5}\] Alternate Solution: Let \[I=\int_{0}^{1}{x{{(1-x)}^{-3/4}}dx}\] \[={{\int_{0}^{1}{(1-x)[1-(1-x)]}}^{-3/4}}dx\] \[=\int_{0}^{1}{(1-x){{x}^{-3/4}}dx}\] \[=\int_{0}^{1}{({{x}^{-3/4}}-{{x}^{1/4}})dx}\] \[=\left[ \frac{{{x}^{1/4}}}{1/4}-\frac{{{x}^{5/4}}}{5/4} \right]_{0}^{1}\] \[=\left[ 4-\frac{4}{5} \right]\] \[=\frac{16}{5}\]
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