Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The value of $\dfrac{{5050\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}dx} }}{{\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}dx} }}$ isA.5040B.5051C.5050D.None of these

Last updated date: 13th Jun 2024
Total views: 403.8k
Views today: 10.03k
Verified
403.8k+ 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.

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.