# Solve this following

Question:

If $\mathrm{I}_{1}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)^{100} \mathrm{dx}$ and $\mathrm{I}_{2}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)^{101} \mathrm{dx}$

such that $\mathrm{I}_{2}=\alpha \mathrm{I}_{1}$ then $\alpha$ equals to

1. $\frac{5050}{5051}$

2. $\frac{5050}{5049}$

3. $\frac{5049}{5050}$

4. $\frac{5051}{5050}$

Correct Option: 1,

Solution:

$\mathrm{I}_{1}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)^{100} \mathrm{dx}$ and $\mathrm{I}_{2}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)^{101} \mathrm{dx}$

and $\mathrm{I}_{1}=\lambda \mathrm{I}_{2}$

$\mathrm{I}_{2}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)^{101} \mathrm{dx}$

$\mathrm{I}_{2}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right)\left(1-\mathrm{x}^{50}\right)^{100} \mathrm{dx}$

$\mathrm{I}_{2}=\int_{0}^{1}\left(1-\mathrm{x}^{50}\right) \mathrm{d} \mathrm{x}-\int_{0}^{1} \mathrm{x}^{50} \cdot\left(1-\mathrm{x}^{50}\right)^{100} \mathrm{dx}$

$I_{2}=I_{1}-\int_{0}^{1} X_{I} \cdot \underbrace{X^{49} \cdot\left(1-X^{50}\right)^{100} d x}_{I I}$

Now apply IBP

$I_{2}=I_{1}-\left[x \int x^{49} \cdot\left(1-x^{50}\right)^{100} d x-\int \frac{d(x)}{d x} \cdot \int \frac{d(x)}{d x} \cdot \int x^{49} \cdot\left(1-x^{50}\right)^{100} d x\right]$

Let $\left(1-x^{50}\right)=t$

$-50 x^{49} d x=d t$

$\mathrm{I}_{2}=\mathrm{I}_{1}-\left[\left.\mathrm{x} \cdot\left(-\frac{1}{50}\right) \frac{\left(1-\mathrm{x}^{50}\right)^{101}}{101}\right|_{\mathrm{x}=0} ^{\mathrm{x}=1}-\int_{0}^{1}\left(-\frac{1}{50}\right) \frac{\left(1-\mathrm{x}^{50}\right)^{101}}{101} \mathrm{dx}\right]$

$\mathrm{I}_{2}=\mathrm{I}_{1}-0-\frac{1}{50} \cdot \frac{1}{101} \cdot \mathrm{I}_{2}=\mathrm{I}_{1}-\frac{1}{5050} \mathrm{I}_{2}$

$\mathrm{I}_{2}+\frac{1}{5050} \mathrm{I}_{2}=\mathrm{I}_{1} \Rightarrow \frac{5051}{5050} \mathrm{I}_{2}=\mathrm{I}_{1}$

$\therefore \alpha=\frac{5050}{5051}$

$\mathrm{I}_{2}=\frac{5050}{5051} \mathrm{I}_{1}$

$\because \mathrm{I}_{2}=\alpha \cdot \mathrm{I}_{1}$