The value of k in N for which the integral I_ | vedclass

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(D) We are given $I_n = \int_0^1 (1 - x^k)^n dx$. Using integration by parts,let $u = (1 - x^k)^n$ and $dv = dx$. Then $du = n(1 - x^k)^{n-1} (-k x^{k

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$7$

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(D) We are given $I_n = \int_0^1 (1 - x^k)^n dx$. Using integration by parts,let $u = (1 - x^k)^n$ and $dv = dx$. Then $du = n(1 - x^k)^{n-1} (-k x^{k-1}) dx$ and $v = x$.$I_n = [x(1 - x^k)^n]_0^1 - \int_0^1 x \cdot n(1 - x^k)^{n-1} (-k x^{k-1}) dx$$I_n = 0 + nk \int_0^1 x^k (1 - x^k)^{n-1} dx$$I_n = nk \int_0^1 (x^k - 1 + 1)(1 - x^k)^{n-1} dx$$I_n = nk \int_0^1 [-(1 - x^k)^n + (1 - x^k)^{n-1}] dx$$I_n = -nk I_n + nk I_{n-1}$$I_n(1 + nk) = nk I_{n-1} \Rightarrow \frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1}$Given $147 I_{20} = 148 I_{21}$,we have $\frac{I_{21}}{I_{20}} = \frac{147}{148}$.Using the recurrence relation for $n = 21$: $\frac{I_{21}}{I_{20}} = \frac{21k}{21k + 1}$.Equating the two: $\frac{21k}{21k + 1} = \frac{147}{148}$.$21k \cdot 148 = 147(21k + 1)$$3108k = 3087k + 147$$21k = 147 \Rightarrow k = 7$.

The value of $k \in N$ for which the integral $I_n = \int_0^1 (1 - x^k)^n dx, n \in N$,satisfies $147 I_{20} = 148 I_{21}$ is :

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