The value of k N for which the integral I n = 0 1 ( 1 - x k ) n ? d x , ? n ? N , satisfies 147 I 20 = 148 I 21 is [2024]

Question

The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} {(1 - x^{k})}^{n} d x, n \in N,$ satisfies $147 I_{20} = 148 I_{21}$ is [2024]

Smart Achievers · Accepted Answer

(2)

$I_{n} = \int_{0}^{1} {(1 - x^{k})}^{n} 1 d x$

By ILATE rule, we have

$I_{n} = {[{(1 - x^{k})}^{n} x]}_{0}^{1} + n k \int_{0}^{1} {(1 - x^{k})}^{n - 1} x^{k} d x$

$= - n k \int_{0}^{1} {(1 - x^{k})}^{n - 1} (1 - x^{k} - 1) d x$

$= - n k [\int_{0}^{1} {(1 - x^{k})}^{n} d x - \int_{0}^{1} {(1 - x^{k})}^{n - 1} d x]$

$= - n k$ $I_{n} + n k I_{n - 1} \Rightarrow I_{n} (1 + n k) = n k$ $I_{n - 1}$

$\Rightarrow \frac{I_{n}}{I_{n - 1}} = \frac{n k}{1 + n k} \Rightarrow \frac{I_{21}}{I_{20}} = \frac{21 k}{1 + 21 k} = \frac{147}{148}$

$\Rightarrow 21 k = 147 \Rightarrow k = 7$

Solution

Q.
The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} {(1 - x^{k})}^{n} d x, n \in N,$ satisfies $147 I_{20} = 148 I_{21}$ is [2024]

1 8

2 7

3 14

4 10

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Solution

Q. The value of k∈N for which the integral In=∫01(1-xk)n dx, n∈N, satisfies 147I20=148I21 is [2024]

1 8

2 7

3 14

4 10

Ans. (2) In=∫01(1-xk)n1 dx By ILATE rule, we have In=[(1-xk)nx]01+nk∫01(1-xk)n-1xk dx =-nk∫01(1-xk)n-1(1-xk-1) dx =-nk[∫01(1-xk)n dx-∫01(1-xk)n-1 dx] =-nk In+nkIn-1⇒In(1+nk)=nk In-1 ⇒InIn-1=nk1+nk⇒I21I20=21k1+21k=147148 ⇒21k=147⇒k=7

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Q.
The value of $k \in N$ for which the integral $I_{n} = \int_{0}^{1} {(1 - x^{k})}^{n} d x, n \in N,$ satisfies $147 I_{20} = 148 I_{21}$ is [2024]