What is \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) equal to?

This question was previously asked in
NDA (Held On: 22 April 2018) Maths Previous Year paper
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  1. 1/110
  2. 1/132
  3. 1/148
  4. 1/140

Answer (Detailed Solution Below)

Option 1 : 1/110
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Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
Calculation:

Let f(x) = x(1 – x)9

Now using property, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)

\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}} = \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){\left\{ {1 - \left( {1 - {\rm{x}}} \right)} \right\}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){{\rm{x}}^9}{\rm{dx}}\)

\(⇒ \mathop \smallint \limits_0^1 \left( {{{\rm{x}}^9} - {{\rm{x}}^{10}}} \right){\rm{dx}}\)

\(⇒\left[ {\frac{{{{\rm{x}}^{10}}}}{{10}} - \frac{{{{\rm{x}}^{11}}}}{{11}}} \right]_0^1\)

⇒ 1/10 – 1/11

1/110

∴ The value of integral \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) is 1/110.

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