JEE PYQ: Indefinite Integration Question 13
Question 13 - 2020 (06 Sep Shift 1)
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:
(1) $\frac{5049}{5050}$
(2) $\frac{5050}{5049}$
(3) $\frac{5050}{5051}$
(4) $\frac{5051}{5050}$
Show Answer
Answer: (3)
Solution
$I_2 = \int_0^1 (1-x^{50})^{100}(1-x^{50}),dx = I_1 - \int_0^1 x^{50}(1-x^{50})^{100},dx$. By parts on the second integral: $\int_0^1 x \cdot x^{49}(1-x^{50})^{100},dx = \left[-\frac{x}{5050}(1-x^{50})^{101}\right]_0^1 + \frac{1}{5050}\int_0^1 (1-x^{50})^{101},dx = \frac{I_2}{5050}$. So $I_2 = I_1 - \frac{I_2}{5050}$, giving $\frac{5051}{5050}I_2 = I_1$, hence $\alpha = \frac{5050}{5051}$.
BETA
