Answer
414.9k+ views
Hint: First we will assume that \[{I_1} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \] and \[{I_2} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} \]. Then we will apply integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[{I_2}\] of the above equation, where \[f = {\left( {1 - {x^{50}}} \right)^{101}}\] and \[g' = 1\]. Then we will simplify to find the required value.
Complete step-by-step answer:
We are given that \[\dfrac{{5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} }}{{\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} }}\].
Let us assume that \[{I_1} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \] and \[{I_2} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} \].
So, we have
\[ \Rightarrow \dfrac{{5050{I_1}}}{{{I_2}}}{\text{ ......eq.(1)}}\]
Within the above difference, applying integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[{I_2}\] of the above equation, where \[f = {\left( {1 - {x^{50}}} \right)^{101}}\] and \[g' = 1\], we get
\[
\Rightarrow {I_2} = \left. {{{\left( {1 - {x^{50}}} \right)}^{101}}x} \right|_0^1 - \int\limits_0^1 {\left( {101} \right){{\left( {1 - {x^{50}}} \right)}^{100}}50 \cdot {x^{49}} \cdot xdx} \\
\Rightarrow {I_2} = \left. {{{\left( {1 - {x^{50}}} \right)}^{101}}x} \right|_0^1 - 50\left( {101} \right)\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = \left( {1 - {1^{50}}} \right) \cdot 1 - \left( {1 - {0^{50}}} \right) \cdot 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = \left( {1 - 1} \right) - \left( {1 - 0} \right) \cdot 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = 0 - 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} + 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \\
\Rightarrow {I_2} = - 5050{I_2} + 5050{I_1} \\
\]
Adding \[5050{I_2}\] in the above equation on both sides, we get
\[
\Rightarrow {I_2} + 5050{I_2} = - 5050{I_2} + 5050{I_1} + 5050{I_2} \\
\Rightarrow {I_2} + 5050{I_2} = 5050{I_1} \\
\Rightarrow 5051{I_2} = 5050{I_1} \\
\]
Dividing the above equation by \[{I_2}\] on both sides, we get
\[
\Rightarrow \dfrac{{5051{I_2}}}{{{I_2}}} = \dfrac{{5050{I_1}}}{{{I_2}}} \\
\Rightarrow 5051 = \dfrac{{5050{I_1}}}{{{I_2}}} \\
\Rightarrow \dfrac{{5050{I_1}}}{{{I_2}}} = 5051 \\
\]
Hence, option B is correct.
Note: We need to know that while finding the value of indefinite integral, we have to add the constant in the final answer or else the answer will be incomplete. We have to be really thorough with the integrations and differentiation of the functions. The key point in this question is to use the integration by parts \[\int {fg'} = fg - \int {f'g} \] to solve this problem. Do not forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it, but this problem can only be solved by parts.
Complete step-by-step answer:
We are given that \[\dfrac{{5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} }}{{\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} }}\].
Let us assume that \[{I_1} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \] and \[{I_2} = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} \].
So, we have
\[ \Rightarrow \dfrac{{5050{I_1}}}{{{I_2}}}{\text{ ......eq.(1)}}\]
Within the above difference, applying integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[{I_2}\] of the above equation, where \[f = {\left( {1 - {x^{50}}} \right)^{101}}\] and \[g' = 1\], we get
\[
\Rightarrow {I_2} = \left. {{{\left( {1 - {x^{50}}} \right)}^{101}}x} \right|_0^1 - \int\limits_0^1 {\left( {101} \right){{\left( {1 - {x^{50}}} \right)}^{100}}50 \cdot {x^{49}} \cdot xdx} \\
\Rightarrow {I_2} = \left. {{{\left( {1 - {x^{50}}} \right)}^{101}}x} \right|_0^1 - 50\left( {101} \right)\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = \left( {1 - {1^{50}}} \right) \cdot 1 - \left( {1 - {0^{50}}} \right) \cdot 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = \left( {1 - 1} \right) - \left( {1 - 0} \right) \cdot 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}{x^{50}}dx} \\
\Rightarrow {I_2} = 0 - 0 - 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} + 5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} \\
\Rightarrow {I_2} = - 5050{I_2} + 5050{I_1} \\
\]
Adding \[5050{I_2}\] in the above equation on both sides, we get
\[
\Rightarrow {I_2} + 5050{I_2} = - 5050{I_2} + 5050{I_1} + 5050{I_2} \\
\Rightarrow {I_2} + 5050{I_2} = 5050{I_1} \\
\Rightarrow 5051{I_2} = 5050{I_1} \\
\]
Dividing the above equation by \[{I_2}\] on both sides, we get
\[
\Rightarrow \dfrac{{5051{I_2}}}{{{I_2}}} = \dfrac{{5050{I_1}}}{{{I_2}}} \\
\Rightarrow 5051 = \dfrac{{5050{I_1}}}{{{I_2}}} \\
\Rightarrow \dfrac{{5050{I_1}}}{{{I_2}}} = 5051 \\
\]
Hence, option B is correct.
Note: We need to know that while finding the value of indefinite integral, we have to add the constant in the final answer or else the answer will be incomplete. We have to be really thorough with the integrations and differentiation of the functions. The key point in this question is to use the integration by parts \[\int {fg'} = fg - \int {f'g} \] to solve this problem. Do not forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it, but this problem can only be solved by parts.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)