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

$\frac{5050}{5051}$

Explanation for correct option

Given: $I_1={\int }_0^1{(1-x^{50})}^{100}dx$ and $I_2={\int }_0^1{(1-x^{50})}^{101}dx$

$$\begin{gathered} I_2={\int }_0^1{(1-x^{50})}^{101}dx \\ \Rightarrow I_2={\int }_0^1{(1-x^{50})(1-x^{50})}^{100}dx \\ \Rightarrow I_2={\int }_0^1{(1-x^{50})}^{100}dx-{\int }_0^1({x^{50}\left)\right(1-x^{50})}^{100}dx \\ \Rightarrow I_2=I_1-{\int }_0^1(x){\left(x^{49}\right)(1-x^{50})}^{100}dx....\left(1\right) \end{gathered}$$

Applying integration by parts i.e. $\int uvdx=u\left(\int vdx\right)-\int u\text{'}\left(\int vdx\right)dx$

$\Rightarrow {\int }_0^1\left(x\right){\left(x^{49}\right)(1-x^{50})}^{100}dx=x{\int }_0^1{\left(x^{49}\right)(1-x^{50})}^{100}dx-{\int }_0^1\left(1\right)\left({\int }_0^1{\left(x^{49}\right)(1-x^{50})}^{100}dx\right)dx$

Let $1-x^{50}=t$

$$\begin{gathered} -50x^{49}dx=dt \\ \Rightarrow x^{49}dx=-\frac{dt}{50} \end{gathered}$$

$\Rightarrow {\int }_0^1\left(x\right){\left(x^{49}\right)(1-x^{50})}^{100}dx=x{\int }_0^1\frac{1}{50}{\int }_0^1{t}^{100}dt-{\int }_0^1\left(1\right)\left({\int }_0^1\frac{1}{50}{\int }_0^1{t}^{100}dt\right)dx$

$$\begin{gathered} ={\left(x\frac{1}{50}\left|\frac{{(1-x^{50})}^{101}}{101}\right|\right)}_0^1-\frac{1}{50}{\int }_0^1\frac{{(1-x^{50})}^{101}}{101}dx \\ =\left(\left(1\right)\frac{1}{50}\left|\frac{{(1-{\left(1\right)}^{50})}^{101}}{101}\right|-0\right)-\frac{1}{5050}{I}_2 \\ =0-\frac{1}{5050}{I}_2 \end{gathered}$$

$\Rightarrow {\int }_0^1\left(x\right){\left(x^{49}\right)(1-x^{50})}^{100}dx=\frac{-1}{5050}{I}_2$

Substituting the above in equation $\left(1\right)$,

$$\begin{gathered} \Rightarrow I_2=I_1-\frac{I_2}{5050} \\ \Rightarrow I_2+\frac{I_2}{5050}=I_1 \\ \Rightarrow \frac{5051}{5050}{I}_2=I_1 \\ \Rightarrow I_2=\frac{5050}{5051}{I}_1 \end{gathered}$$

Now comparing it with $I_2=\alpha I_1$$\Rightarrow \alpha =\frac{5050}{5051}$.

Hence, option(B) i.e. $\frac{5050}{5051}$ is correct.