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Question

Mathematics Question on Definite Integral

The value of the integral ∫01x(1−x)49 dx\int_{0}^{1}{x{{(1-x)}^{49}}\,dx} is equal to

A

12550\frac{1}{2550}

B

12500\frac{1}{2500}

C

10490\frac{10}{490}

D

149\frac{1}{49}

Answer

12550\frac{1}{2550}

Explanation

Solution

Let I=∫01x(1−x)49dxI=\int_{0}^{1}{x{{(1-x)}^{49}}dx}
=∫01(1−x) [1−(1−x)]49 dx=\int_{0}^{1}{(1-x)\,{{[1-(1-x)]}^{49}}\,dx}
[∵  ∫0af(x) dx=∫0af(a−x) dx]\left[ \because \,\,\int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \right]
=∫01(1−x)x49 dx=\int_{0}^{1}{(1-x){{x}^{49}}\,dx}
=∫01(x49−x50)dx=\int_{0}^{1}{({{x}^{49}}-{{x}^{50}})dx}
=[x5050−x5151]01=\left[ \frac{{{x}^{50}}}{50}-\frac{{{x}^{51}}}{51} \right]_{0}^{1}
=(150−151)−0=150×51=12550=\left( \frac{1}{50}-\frac{1}{51} \right)-0=\frac{1}{50\times 51}=\frac{1}{2550}