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 $1-x=t\Rightarrow -dx=dt$ $\therefore$ $I={\int }_1^0(1-t)t^{99}dt$ $={\int }_0^1[t^{99}-t^{100}]dt={\left[\frac{t^{100}}{100}-\frac{t^{101}}{101}\right]}_0^1$ $=\frac{1}{100}-\frac{1}{101}=\frac{1}{10100}$