What is the value of the integral \[I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} \]?
A. \[\dfrac{1}{{n + 1}}\]
B. \[\dfrac{1}{{n + 2}}\]
C. \[\dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}\]
D. \[\dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}}\]

Hint: Here, a definite integral is given. First, multiply the given integral by \[ - 1\] and simplify it. Then, rewrite \[\left( { - x} \right)\] as \[\left( {1 - x - 1} \right)\] in the integral. After that, perform some mathematical operations and simplify the integral. Then, apply the subtraction rule of the integration and solve the integrals by using the standard integral formula. In the end, apply the upper and the lower limits to get the required answer.

Formula Used: \[\int\limits_a^b {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int\limits_a^b {f\left( x \right)dx \pm } \int\limits_a^b {g\left( x \right)dx} \]
\[\int\limits_a^b {{{\left( {1 - x} \right)}^n}dx} = \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^b\]

Complete step by step solution: The given definite integral is \[I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} \].

Let’s simplify the given interval.
Multiply both sides by \[ - 1\].
\[ - I = \int\limits_0^1 { - x{{\left( {1 - x} \right)}^n}dx} \]
Rewrite \[\left( { - x} \right)\] by adding and subtracting 1.
\[ - I = \int\limits_0^1 {\left( {1 - x - 1} \right){{\left( {1 - x} \right)}^n}dx} \]
Simplify the right-hand side.
\[ - I = \int\limits_0^1 {\left[ {\left( {1 - x} \right){{\left( {1 - x} \right)}^n} - 1{{\left( {1 - x} \right)}^n}} \right]dx} \]
\[ \Rightarrow - I = \int\limits_0^1 {\left[ {{{\left( {1 - x} \right)}^{n + 1}} - {{\left( {1 - x} \right)}^n}} \right]dx} \]
Now apply the subtraction rule of the integration on the right-hand side.
\[ \Rightarrow - I = \int\limits_0^1 {{{\left( {1 - x} \right)}^{n + 1}}dx - } \int\limits_0^1 {{{\left( {1 - x} \right)}^n}dx} \]
Solve the integrals by using the integration formula \[\int\limits_a^b {{{\left( {1 - x} \right)}^n}dx} = \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^b\].
We get,
\[ \Rightarrow - I = \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1 + 1}}}}{{n + 1 + 1}}} \right]_a^1 - \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^1\]
\[ \Rightarrow - I = \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 2}}}}{{n + 2}}} \right]_a^1 + \left[ {\dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^1\]
Apply the upper and lower limit on the right-hand side.
\[ \Rightarrow - I = \left[ { - \dfrac{{{{\left( {1 - 1} \right)}^{n + 2}}}}{{n + 2}} + \dfrac{{{{\left( {1 - 0} \right)}^{n + 2}}}}{{n + 2}}} \right] + \left[ {\dfrac{{{{\left( {1 - 1} \right)}^{n + 1}}}}{{n + 1}} - \dfrac{{{{\left( {1 - 0} \right)}^{n + 1}}}}{{n + 1}}} \right]\]
\[ \Rightarrow - I = \left[ { - \dfrac{{{{\left( 0 \right)}^{n + 2}}}}{{n + 2}} + \dfrac{{{{\left( 1 \right)}^{n + 2}}}}{{n + 2}}} \right] + \left[ {\dfrac{{{{\left( 0 \right)}^{n + 1}}}}{{n + 1}} - \dfrac{{{{\left( 1 \right)}^{n + 1}}}}{{n + 1}}} \right]\]
\[ \Rightarrow - I = \left[ {\dfrac{1}{{n + 2}}} \right] + \left[ { - \dfrac{1}{{n + 1}}} \right]\]
\[ \Rightarrow - I = \dfrac{1}{{n + 2}} - \dfrac{1}{{n + 1}}\]
Again, multiply both sides by \[ - 1\].
\[ \Rightarrow I = \dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}\]

Option ‘C’ is correct

Note: Sometimes students get confused and solve the above given integral by using the power rule of integration. Then get \[\int\limits_a^b {{{\left( {1 - x} \right)}^n}dx} = \left[ {\dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^b\], which is incorrect formula. The correct formula is \[\int\limits_a^b {{{\left( {1 - x} \right)}^n}dx} = \left[ { - \dfrac{{{{\left( {1 - x} \right)}^{n + 1}}}}{{n + 1}}} \right]_a^b\].