If I_{1} = int_{0}^{1} (1 - x^{50})^{100} dx | vedclass

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(A) We are given $I_{1} = \int_{0}^{1} (1 - x^{50})^{100} dx$ and $I_{2} = \int_{0}^{1} (1 - x^{50})^{101} dx$. We can write $I_{2}$ as: $I_{2} = \int

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$\frac{5050}{5051}$

Vedclass · Answer

(A) We are given $I_{1} = \int_{0}^{1} (1 - x^{50})^{100} dx$ and $I_{2} = \int_{0}^{1} (1 - x^{50})^{101} dx$. We can write $I_{2}$ as: $I_{2} = \int_{0}^{1} (1 - x^{50}) (1 - x^{50})^{100} dx$ $I_{2} = \int_{0}^{1} (1 - x^{50})^{100} dx - \int_{0}^{1} x^{50} (1 - x^{50})^{100} dx$ $I_{2} = I_{1} - \int_{0}^{1} x \cdot (x^{49} (1 - x^{50})^{100}) dx$ Using Integration by Parts $(IBP)$ for the second integral,let $u = x$ and $dv = x^{49} (1 - x^{50})^{100} dx$. Then $du = dx$ and $v = \int x^{49} (1 - x^{50})^{100} dx$. Let $1 - x^{50} = t$,then $-50x^{49} dx = dt$,so $x^{49} dx = -\frac{dt}{50}$. $v = \int t^{100} (-\frac{dt}{50}) = -\frac{t^{101}}{5050} = -\frac{(1 - x^{50})^{101}}{5050}$. Applying $IBP$: $\int_{0}^{1} x \cdot (x^{49} (1 - x^{50})^{100}) dx = [x \cdot (-\frac{(1 - x^{50})^{101}}{5050})]_{0}^{1} - \int_{0}^{1} (-\frac{(1 - x^{50})^{101}}{5050}) dx$ $= [0 - 0] + \frac{1}{5050} \int_{0}^{1} (1 - x^{50})^{101} dx = \frac{1}{5050} I_{2}$. Substituting this back into the equation for $I_{2}$: $I_{2} = I_{1} - \frac{1}{5050} I_{2}$ $I_{2} (1 + \frac{1}{5050}) = I_{1}$ $I_{2} (\frac{5051}{5050}) = I_{1}$ $I_{2} = \frac{5050}{5051} I_{1}$. Since $I_{2} = \alpha I_{1}$,we have $\alpha = \frac{5050}{5051}$.

If $I_{1} = \int_{0}^{1} (1 - x^{50})^{100} dx$ and $I_{2} = \int_{0}^{1} (1 - x^{50})^{101} dx$ such that $I_{2} = \alpha I_{1}$,then $\alpha$ is equal to:

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