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Question: The value of \[\dfrac{{(5050)\int_0^1 {(1 - {x^{50}}} {)^{100}}dx}}{{\int_0^1 {{{(1 - {x^{50}})}^{10...

The value of (5050)∫01(1−x50)100dx∫01(1−x50)101dx\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

Explanation

Solution

We try to generalize this problem in our way to solve it easily. We start with In+1{I_{n + 1}}and In{I_n}in general as terms, In=∫01(1−xm)ndx{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx. And similarly, we’ll get the value of In+1{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: (5050)∫01(1−x50)100dx∫01(1−x50)101dx\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, In=∫01(1−xm)ndx{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx
Substituting n=n+1
Now, breaking the exponent asax+y=axay{a^{x + y}} = {a^x}{a^y}
=∫01(1−xm)n(1−xm)dx= \int_0^1 {{{(1 - {x^m})}^n}(1 - {x^m})} dx
On multiplying we get,
=∫01[(1−xm)n−xm(1−xm)n]dx= \int_0^1 {[{{(1 - {x^m})}^n} - {x^m}{{(1 - {x^m})}^n}} ]dx
As we know that∫(A+B)dx=∫Adx+∫Bdx\int {(A + B)dx} = \int {Adx} + \int {Bdx}
=∫01(1−xm)ndx−∫01xm(1−xm)ndx= \int_0^1 {{{(1 - {x^m})}^n}dx - \int_0^1 {{x^m}{{(1 - {x^m})}^n}} } dx
Now, from ∫01(1−xm)ndx=In\int_0^1 {{{(1 - {x^m})}^n}} dx = {I_n},
=In−∫01xm(1−xm)ndx= {I_n} - \int_0^1 {{x^m}{{(1 - {x^m})}^n}dx}
=In−∫01x.xm−1(1−xm)ndx= {I_n} - \int_0^1 {x.} {x^{m - 1}}{(1 - {x^m})^n}dx
By integrating by parts,
i.e. ∫f(x)g(x)dx=f(x)∫g(x)dx−∫(f′(x)∫g(x)dx)dx+c\int {f(x)g(x)dx} = f(x)\int {g(x)dx} - \int {(f'(x)\int {g(x)dx} )dx + c}
=In−[−x(1−xm)n+1m(n+1)∣01+∫01(1−xm)n+1m(n+1)dx]= {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]}
=In+[1(1−1m)n+1−0(1−0m)n+1m(n+1)−1m(n+1)∫01(1−xm)n+1dx]= {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 =In+1 = {I_{n + 1}}
Now, we have,
In+1=In−1m(n+1)In+1{I_{n + 1}} = {I_n} - \dfrac{1}{{m(n + 1)}}{I_{n + 1}}
On changing sides, we get,
In+1+1m(n+1)In+1=In{I_{n + 1}} + \dfrac{1}{{m(n + 1)}}{I_{n + 1}} = {I_n}
⇒In+1[m(n+1)+1]m(n+1)=In\Rightarrow \dfrac{{{I_{n + 1}}[m(n + 1) + 1]}}{{m(n + 1)}} = {I_n}
On cross multiplying, we get,
⇒In+1[m(n+1)+1]=Inm(n+1)\Rightarrow {I_{n + 1}}[m(n + 1) + 1] = {I_n}m(n + 1)
Now, in our problem, n = 100,m = 50{\text{n = 100,m = 50}}
So we get,
I100+1[50(100+1)+1]=I10050(100+1){I_{100 + 1}}[50(100 + 1) + 1] = {I_{100}}50(100 + 1)
⇒I101[50(101)+1]=I10050(101)\Rightarrow {I_{101}}[50(101) + 1] = {I_{100}}50(101)
⇒[50(101)+1]=I1005050I101\Rightarrow [50(101) + 1] = \dfrac{{{I_{100}}5050}}{{{I_{101}}}}
⇒I1005050I101=5051\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 In+1{I_{n + 1}} to find the form In{I_n}to reach our answer.In+1{I_{n + 1}}is found by the same formula In{I_n}, we can also proceed with the problem with In{I_n}and In−1{I_{n - 1}} if we consider it n=101n = 101.
Here we have found that if In=∫01(1−xm)ndx{I_n} = \int_0^1 {{{(1 - {x^m})}^n}} dx then In+1[m(n+1)+1]=Inm(n+1){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.