Question

The value of \[\dfrac{{(5050)\int_0^1 {(1 - {x^{50}}} {)^{100}}dx}}{{\int_0^1 {{{(1 - {x^{50}})}^{101}}dx} }}\]is,
A. 5040
B. 5051
C. 5050
D. None of these

Answer 1

Hint: We try to generalize this problem in our way to solve it easily. We start with \[{I_{n + 1}}\]and \[{I_n}\]in general as terms, \[{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx\]. And similarly, we’ll get the value of \[{I_{n + 1}}\] . So on comparing the term with the term given the question we’ll substitute the value of n to get the required answer.

Complete step by step Answer:

Given: \[\dfrac{{(5050)\int_0^1 {(1 - {x^{50}}} {)^{100}}dx}}{{\int_0^1 {{{(1 - {x^{50}})}^{101}}dx} }}\], we need to find its value.
Let, \[{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx\]
Substituting n=n+1
Now, breaking the exponent as${a^{x + y}} = {a^x}{a^y}$
\[ = \int_0^1 {{{(1 - {x^m})}^n}(1 - {x^m})} dx\]
On multiplying we get,
\[ = \int_0^1 {[{{(1 - {x^m})}^n} - {x^m}{{(1 - {x^m})}^n}} ]dx\]
As we know that$\int {(A + B)dx} = \int {Adx} + \int {Bdx} $
\[ = \int_0^1 {{{(1 - {x^m})}^n}dx - \int_0^1 {{x^m}{{(1 - {x^m})}^n}} } dx\]
Now, from \[\int_0^1 {{{(1 - {x^m})}^n}} dx = {I_n}\],
\[ = {I_n} - \int_0^1 {{x^m}{{(1 - {x^m})}^n}dx} \]
\[ = {I_n} - \int_0^1 {x.} {x^{m - 1}}{(1 - {x^m})^n}dx\]
By integrating by parts,
i.e. $\int {f(x)g(x)dx} = f(x)\int {g(x)dx} - \int {(f'(x)\int {g(x)dx} )dx + c} $
\[ = {I_n} - [\left. { - \dfrac{{x{{(1 - {x^m})}^{n + 1}}}}{{m(n + 1)}}} \right|_0^1 + \int_0^1 {\dfrac{{{{(1 - {x^m})}^{n + 1}}}}{{m(n + 1)}}dx]} \]
\[ = {I_n} + [\dfrac{{1{{(1 - {1^m})}^{n + 1}} - 0{{(1 - {0^m})}^{n + 1}}}}{{m(n + 1)}} - \dfrac{1}{{m(n + 1)}}\int_0^1 {{{(1 - {x^m})}^{n + 1}}dx]} \]
Our first term turns out to be zero and the 2nd term is \[ = {I_{n + 1}}\]
Now, we have,
\[{I_{n + 1}} = {I_n} - \dfrac{1}{{m(n + 1)}}{I_{n + 1}}\]
On changing sides, we get,
\[{I_{n + 1}} + \dfrac{1}{{m(n + 1)}}{I_{n + 1}} = {I_n}\]
\[ \Rightarrow \dfrac{{{I_{n + 1}}[m(n + 1) + 1]}}{{m(n + 1)}} = {I_n}\]
On cross multiplying, we get,
\[ \Rightarrow {I_{n + 1}}[m(n + 1) + 1] = {I_n}m(n + 1)\]
Now, in our problem, \[{\text{n = 100,m = 50}}\]
So we get,
\[{I_{100 + 1}}[50(100 + 1) + 1] = {I_{100}}50(100 + 1)\]
\[ \Rightarrow {I_{101}}[50(101) + 1] = {I_{100}}50(101)\]
\[ \Rightarrow [50(101) + 1] = \dfrac{{{I_{100}}5050}}{{{I_{101}}}}\]
\[ \Rightarrow \dfrac{{{I_{100}}5050}}{{{I_{101}}}} = 5051\]
So, we have our answer as, 5051, which is an option (b)

Note: Here to solve the problem we have used some formulas of integration. We calculated \[{I_{n + 1}}\] to find the form \[{I_n}\]to reach our answer.\[{I_{n + 1}}\]is found by the same formula \[{I_n}\], we can also proceed with the problem with \[{I_n}\]and \[{I_{n - 1}}\] if we consider it \[n = 101\].
Here we have found that if \[{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx\] then \[{I_{n + 1}}[m(n + 1) + 1] = {I_n}m(n + 1)\]. Therefore, we found the generalized term for this type of question to easily simplify to get the required answers.