Question

The value of $ \int_{0}^{1}{x\,{{(1-x)}^{99}}}\,dx $ is equal to:

• A. $ \frac{1}{10100} $
• B. $ \frac{11}{10100} $
• C. $ \frac{1}{10010} $
• D. $ \frac{11}{11100} $
• E. $ \frac{1}{1010} $

Answer 1

Answer: (A)

Let $ I=\int_{0}^{1}{x}{{(1-x)}^{99}}dx $ Put $ 1-x=t\Rightarrow -dx=dt $ $ \therefore $ $ I=\int_{1}^{0}{(1-t)}{{t}^{99}}dt $ $ =\int_{0}^{1}{[{{t}^{99}}-{{t}^{100}}]}dt=\left[ \frac{{{t}^{100}}}{100}-\frac{{{t}^{101}}}{101} \right]_{0}^{1} $ $ =\frac{1}{100}-\frac{1}{101}=\frac{1}{10100} $