If -01200 C 6 x 1941- x 6 dx -1- k- where k - N- then the value of k is

I=^200C_6∫ ^1_0 x^194(1 x)^6dx Using by parts I=^200C_6[[ nbsp; (1 x)^6×x^195195]_0^1+6195[ ∫_0^1(1 x)^5× x^195dx]] Here, [ nbsp; (1 x)^6×x^195195]_0^1=0 ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

If ∫01200 C 6...

Question

If $\int_{0}^{1}^{200} C_{6} x^{194} (1 - x)^{6} d x = \frac{1}{k}$ , where $k \in N$ , then the value of $k$ is

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Solution

$I =^{200} C_{6} \int_{0}^{1} x^{194} (1 - x)^{6} d x$
Using by parts -
$I =^{200} C_{6} [{[(1 - x)^{6} \times \frac{x^{195}}{195}]}_{0}^{6} + \frac{6}{195} [\int_{0}^{1} (1 - x)^{5} \times x^{195} d x]]$
Here, ${[(1 - x)^{6} \times \frac{x^{195}}{195}]}_{0}^{6} = 0$
$∴ I =^{200} C_{6} \frac{6}{195} [\int_{0}^{1} (1 - x)^{5} \times x^{195} d x]$
Integrating by parts again $5$ more times, we get -
$I =^{200} C_{6} \frac{6!}{195 \times 196 \times 197 \times 198 \times 199 \times 200} \times \int_{0}^{1} x^{200} d x I = {[\frac{x^{201}}{201}]}_{0}^{1} = \frac{1}{201} \Rightarrow k = 201$

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If ∫10200C6x194(1−x)6dx=1k , where k∈N , then the value of k is

I=200C6∫10x194(1−x)6dx Using by parts - I=200C6[[(1−x)6×x195195]10+6195[∫10(1−x)5×x195dx]] Here, [(1−x)6×x195195]10=0 ∴I= 200C66195[∫10(1−x)5×x195dx] Integrating by parts again 5 more times, we get - I= 200C66!195×196×197×198×199×200×∫10x200dxI=[x201201]10=1201⇒k=201

If $\int_{0}^{1}^{200} C_{6} x^{194} (1 - x)^{6} d x = \frac{1}{k}$ , where $k \in N$ , then the value of $k$ is