Question

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 $\alpha$ equals to:

• A. $\frac{5050}{5049}$
• B. $\frac{5050}{5051}$
• C. $\frac{5051}{5050}$
• D. $\frac{5049}{5050}$

Answer 1

The correct option is B $\frac{5050}{5051}$

$I_1={\int }_0^1(1-x^{50}{)}^{100}dx$
$I_2={\int }_0^1(1-x^{50})(1-x^{50}{)}^{100}dx$
$={\int }_0^1(1-x^{50}{)}^{100}dx-{\int }_0^1{x}^{50}(1-x^{50}{)}^{100}dx$
$I_2=I_1-{\int }_0^1\underset{I}{\underset{}{x}}\cdot \underset{II}{\underset{}{x^{49}{(1-x^{50})}^{100}}}dx$
By using by parts
$1-x^{50}=t$
$\Rightarrow x^{49}dx=\frac{-dt}{50}$
$I_2=I_1-{\left[x\left(\frac{-1}{50}\right)\frac{(1-x^{50}{)}^{101}}{101}\right]}_0^1+{\int }_1^0\left(\frac{-1}{50}\right)\frac{(1-x^{50}{)}^{101}}{101}dx$
$I_2=I_1-0+\frac{{\int }_0^1(1-x^{50}{)}^{101}}{(-5050)}$
$I_2=I_1-\frac{I_2}{5050}$
$\frac{5051}{5050}{I}_2=I_1$
$I_2=\frac{5050}{5051}{I}_1$
$\alpha =\frac{5050}{5051}$