Question
Download Solution PDF\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
निश्चित समाकल गुण:
\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
गणना:
माना कि f(x) = x(1 – x)9
अब गुण का प्रयोग करने पर, \(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {{\rm{a}} + {\rm{b}} - {\rm{x}}} \right){\rm{\;dx}}\)
\(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}} = \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){\left\{ {1 - \left( {1 - {\rm{x}}} \right)} \right\}^9}{\rm{dx}}\)
\(= \mathop \smallint \limits_0^1 \left( {1 - {\rm{x}}} \right){{\rm{x}}^9}{\rm{dx}}\)
\(= \mathop \smallint \limits_0^1 \left( {{{\rm{x}}^9} - {{\rm{x}}^{10}}} \right){\rm{dx}}\)
\(= \left[ {\frac{{{{\rm{x}}^{10}}}}{{10}} - \frac{{{{\rm{x}}^{11}}}}{{11}}} \right]_0^1\)
= 1/10 – 1/11
= 1/110
∴ समाकलन \(\mathop \smallint \nolimits_0^1 {\rm{x}}{(1 - {\rm{x}})^9}{\rm{dx}}\) का मान 1/110 है।
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