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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Hard
1 of 1
Application +4 | -1
For each positive integer nn, define a function fnf_n on [0,1][0,1] as follows:

fn(x)={0ifx=0sin⁡π2nif0<x≤1nsin⁡2π2nif1n<x≤2nsin⁡3π2nif2n<x≤3nsin⁡nπ2nifn−1n<x≤1f_n(x)=\left\{\begin{array}{ccc} 0 & \text {if} & x=0 \\ \sin \frac{\pi}{2 n} & \text {if} & 0 < x \leq \frac{1}{n} \\ \sin \frac{2 \pi}{2 n} & \text {if} & \frac{1}{n} < x \leq \frac{2}{n} \\ \sin \frac{3 \pi}{2 n} & \text {if} & \frac{2}{n} < x \leq \frac{3}{n} \\ \sin \frac{n \pi}{2 n} & \text {if} & \frac{n-1}{n} < x \leq 1 \end{array}\right.

Then the value of lim⁡n→∞∫01fn(x)dx\lim \limits_{{n \rightarrow \infty}} \int\limits_{0}^1 f_n(x) d x is:
View answer and explanation
Correct option: D
2π\frac{2}{\pi}
Option D (2π\frac{2}{\pi}) is correct because the limit of the integral is evaluated by recognizing it as a Riemann sum and converting it to a definite integral.

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