If I1 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ and
I2 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ such
that I2
= $\alpha$I1
then $\alpha$
equals to :
Solution
I<sub>2</sub> = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$
<br><br>I<sub>2</sub> = $$\int\limits_0^1 {\left( {1 - {x^{50}}} \right){{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
<br><br>I<sub>2</sub> = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} $$
<br><br>I<sub>2</sub> = I<sub>1</sub> - $\int\limits_0^1 {x.{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx}$
<br><br>Now apply IBP
<br><br>I<sub>2</sub> = I<sub>1</sub> - <br>$$\left[ {x\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} - \int {{{d\left( x \right)} \over {dx}}.\int {{{d\left( x \right)} \over {dx}}\int\limits_0^1 {{x^{49}}{{\left( {1 - {x^{50}}} \right)}^{100}}dx} } } } \right]$$
<br><br>Let (1 – x<sup>50</sup>) = t
<br><br>$\Rightarrow$ -50x<sup>49</sup>dx = dt
<br><br>I<sub>2</sub> = I<sub>1</sub> - <br>$$\left[ {x.\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}} \right]_0^1$$ - $$ - \int\limits_0^1 {\left( { - {1 \over {50}}} \right){{{{\left( {1 - {x^{50}}} \right)}^{101}}} \over {101}}dx} $$
<br><br>= I<sub>1</sub> - 0 - ${ - {1 \over {50}}.{1 \over {101}}{I_2}}$
<br><br>I<sub>2</sub> = I<sub>1</sub> - ${1 \over {5050}}{I_2}$
<br><br>$\Rightarrow$ I<sub>2</sub> + ${1 \over {5050}}{I_2}$ = I<sub>1</sub>
<br><br>$\Rightarrow$ ${{5051} \over {5050}}{I_2}$ = I<sub>1</sub>
<br><br>$\Rightarrow$ I<sub>2</sub> = ${{5050} \over {5051}}$I<sub>1</sub>
<br><br>Given I<sub>2</sub>
= $\alpha$I<sub>1</sub>
<br><br>$\therefore$ $\alpha$ = ${{5050} \over {5051}}$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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