Question

The value of the integral ${\int }_0^{1}x{(1-x)}^{49}dx$ is equal to

• A. $\frac{1}{2550}$
• B. $\frac{1}{2500}$
• C. $\frac{10}{490}$
• D. $\frac{1}{49}$

Answer 1

Answer: (A)

Let $I={\int }_0^{1}x{(1-x)}^{49}dx$
$={\int }_0^1(1-x){[1-(1-x)]}^{49}dx$
$\left[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx\right]$
$={\int }_0^1(1-x)x^{49}dx$
$={\int }_0^1(x^{49}-x^{50})dx$
$={\left[\frac{x^{50}}{50}-\frac{x^{51}}{51}\right]}_0^1$
$=\left(\frac{1}{50}-\frac{1}{51}\right)-0=\frac{1}{50\times 51}=\frac{1}{2550}$