Question

If $I_1=\underset{0}{\overset{1}{\int }}{(1-x^{50})}^{100}dx,I_2=\underset{0}{\overset{1}{\int }}{(1-x^{50})}^{101}dx$, then

• A. $51I_1=50I_2$
• B. $101I_1=100I_2$
• C. $5051I_1=5050I_2$
• D. $5050I_1=5051I_2$

Answer 1

The correct option is D $5050I_1=5051I_2$

Given: $I_1=\underset{0}{\overset{1}{\int }}{(1-x^{50})}^{100}dx{I}_2=\underset{0}{\overset{1}{\int }}{(1-x^{50})}^{101}dx$

Now,
$I_2=\underset{0}{\overset{1}{\int }}(1-x^{50}){(1-x^{50})}^{100}dx\Rightarrow I_2=\underset{0}{\overset{1}{\int }}{(1-x^{50})}^{100}dx-\underset{0}{\overset{1}{\int }}{x}^{50}{(1-x^{50})}^{100}dx\Rightarrow I_2=I_1-\underset{0}{\overset{1}{\int }}x\cdot x^{49}{(1-x^{50})}^{100}dx\Rightarrow I_2=I_1-I$
Now,
$I=\underset{0}{\overset{1}{\int }}x\cdot x^{49}{(1-x^{50})}^{100}dx\Rightarrow I=x\underset{0}{\overset{1}{\int }}{x}^{49}{(1-x^{50})}^{100}dx-\underset{0}{\overset{1}{\int }}\left(\frac{d\left(x\right)}{dx}\int x^{49}{(1-x^{50})}^{100}\right)dx$
Taking
$(1-x^{50})=t\Rightarrow x^{49}dx=-\frac{dt}{50}\Rightarrow I={[-\frac{x(1-x^{50}{)}^{101}}{50\times 101}]}_0^1+\frac{1}{50\times 101}\underset{0}{\overset{1}{\int }}(1-x^{50}{)}^{101}dx\Rightarrow I=\frac{I_2}{5050}$
Therefore,
$I_2=I_1-\frac{I_2}{5050}\therefore 5051I_2=5050I_1$