If n(2n+1) int_{0}^{1}(1-x^n)^{2n} dx = 1177 | vedclass

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(B) Let $I_1 = \int_{0}^{1}(1-x^n)^{2n} dx$ and $I_2 = \int_{0}^{1}(1-x^n)^{2n+1} dx$.Using integration by parts for $I_2$:$I_2 = \int_{0}^{1} (1-x^n)

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Let $I_1 = \int_{0}^{1}(1-x^n)^{2n} dx$ and $I_2 = \int_{0}^{1}(1-x^n)^{2n+1} dx$.Using integration by parts for $I_2$:$I_2 = \int_{0}^{1} (1-x^n)^{2n+1} \cdot 1 dx$$= [x(1-x^n)^{2n+1}]_0^1 - \int_{0}^{1} x \cdot (2n+1)(1-x^n)^{2n} \cdot (-nx^{n-1}) dx$$= 0 - (2n+1)(-n) \int_{0}^{1} x^n (1-x^n)^{2n} dx$$= n(2n+1) \int_{0}^{1} (x^n - 1 + 1)(1-x^n)^{2n} dx$$= n(2n+1) [\int_{0}^{1} -(1-x^n)^{2n+1} dx + \int_{0}^{1} (1-x^n)^{2n} dx]$$I_2 = n(2n+1) [-I_2 + I_1]$$I_2 = -n(2n+1)I_2 + n(2n+1)I_1$$I_2(1 + 2n^2 + n) = n(2n+1)I_1$$\frac{I_1}{I_2} = \frac{2n^2 + n + 1}{n(2n+1)}$.Given $\frac{I_1}{I_2} = \frac{1177}{n(2n+1)}$,we have $2n^2 + n + 1 = 1177$.$2n^2 + n - 1176 = 0$.Solving the quadratic equation: $n = \frac{-1 \pm \sqrt{1 - 4(2)(-1176)}}{4} = \frac{-1 \pm \sqrt{1 + 9408}}{4} = \frac{-1 \pm 97}{4}$.Since $n \in N$,$n = \frac{96}{4} = 24$.

If $n(2n+1) \int_{0}^{1}(1-x^n)^{2n} dx = 1177 \int_{0}^{1}(1-x^n)^{2n+1} dx$,then $n \in N$ is equal to $\dots\dots$

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