Question

The value of the integral, $I={{\int_{0}^{1}{x\left( 1-x \right)}}^{n}}dx$ is?
(a) $\dfrac{1}{n+1}+\dfrac{1}{n+2}$
(b) $\dfrac{1}{n+1}$
(c) $\dfrac{1}{n+2}$
(d) $\dfrac{1}{n+1}-\dfrac{1}{n+2}$

Answer 1

Hint: Assume the integral be equal to ‘$I$’. Then use the property of definite integral given by: $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$ to simplify the integral and then find the integral value by using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ and substitute the proper limits.

Complete step by step answer:

Here, we have been provided with a definite integral. There are certain properties of definite integral but here we will use a basic property which is, $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$.
Now, let us come to the question. Let us assume the given integral be ‘$I$’. Therefore,
\[I=\int_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}\]
 Now, using the property, $\int\limits_{a}^{b}{f(x)}dx=\int\limits_{a}^{b}{f(a+b-x)}dx$, we get,
\[\begin{align}
  & I=\int_{0}^{1}{\left( 1-x \right){{\left( 1-\left( 1-x \right) \right)}^{n}}dx} \\
 & =\int_{0}^{1}{\left( 1-x \right){{x}^{n}}dx} \\
 & =\int_{0}^{1}{\left( {{x}^{n}}-{{x}^{n+1}} \right)dx} \\
\end{align}\]
Now, breaking the integral into two parts, we get,
\[I=\int_{0}^{1}{{{x}^{n}}dx}-\int_{0}^{1}{{{x}^{n+1}}dx}\]
Using the formula, $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, we get,
$I=\left[ \dfrac{{{x}^{n+1}}}{n+1}-\dfrac{{{x}^{n+2}}}{n+2} \right]_{0}^{1}$
Substituting the limits we get,
$\begin{align}
  & I=\left[ \dfrac{1}{n+1}-\dfrac{1}{n+2} \right]-\left[ \dfrac{0}{n+1}-\dfrac{0}{n+2} \right] \\
 & =\dfrac{1}{n+1}-\dfrac{1}{n+2} \\
\end{align}$
Hence, option (d) is the correct answer.

Note: Properties of definite integrals are very important. Here, we have used one of the properties of definite integral and it became so easy to simplify. If we will not use properties of definite integral here and solve it like an indefinite integral then it will be a very lengthy and time consuming process. So, basic properties of definite integral are important to solve this question.