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Definite Integration: Positive Integers Let Denote Area Aob Origin Aob

JEE Maths question with a full step-by-step solution.

Question

For positive integers

k = 1,2,3,\dots,n

, let

S_k

denote the area of

\triangle AOB_k

(where

O

is the origin) such that

\angle AOB_k = \dfrac{k\pi}{2n}

OA = 1

and

OB_k = k

. Then the value of

\displaystyle\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^{n}S_k

\dfrac2{\pi^2}

correct

\dfrac4{\pi^2}

\dfrac8{\pi^2}

\dfrac1{2\pi^2}

Solution

Step 1: For a triangle with two sides

p

q

and included angle

\theta

the area is

\tfrac12pq\sin\theta

, so

S_k = \frac12\cdot1\cdot k\cdot\sin\left(\frac{k\pi}{2n}\right) = \frac k2\sin\left(\frac{k\pi}{2n}\right)

Step 2:

\frac1{n^2}\sum_{k=1}^{n}S_k = \frac1{n^2}\sum_{k=1}^{n}\frac k2\sin\left(\frac{k\pi}{2n}\right) = \frac12\cdot\frac1n\sum_{k=1}^{n}\frac kn\sin\left(\frac\pi2\cdot\frac kn\right)

Step 3: With

x = \dfrac kn

and mesh

\dfrac1n

\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}f\!\left(\frac kn\right) = \int_0^1f(x)\,dx , \qquad f(x) = x\sin\frac{\pi x}2

so the limit equals

\frac12\int_0^1 x\sin\frac{\pi x}2\,dx

Step 4: Integrating by parts, with

a = \dfrac\pi2

\int x\sin\left(ax\right)dx = -\frac{x\cos ax}{a}+\frac{\sin ax}{a^2}

\int_0^1x\sin\frac{\pi x}2\,dx = \left[-\frac{x\cos\frac{\pi x}2}{\pi/2} +\frac{\sin\frac{\pi x}2}{\left(\pi/2\right)^2}\right]_0^1 = 0+\frac{1}{\pi^2/4} = \frac4{\pi^2}

\frac12\cdot\frac4{\pi^2} = \frac2{\pi^2}

Answer: (1)

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