JEE Main 2022 — Definite Integration Question with Solution

$$\underset{n\to \alpha }{\lim }\sum _{r=1}^n\frac{r}{2r^2-7rn+6n^2}$$

$$=\underset{n\to \alpha }{\lim }\frac{1}{n}\sum _{r=1}^n\frac{\left(\frac{r}{n}\right)}{2{\left(\frac{r}{n}\right)}^2-7\left(\frac{r}{n}\right)+6}$$

$$={\int }_a^{b}f(x)dx$$

$$a=\underset{n\to \alpha }{\lim }\left(\frac{1}{n}\right)=0$$

$$b=\underset{n\to \alpha }{\lim }\left(\frac{n}{n}\right)=1$$

and $$\frac{r}{n}\to x$$

$$={\int }_0^1\frac{x}{2x^2-7x+6}dx$$

$$={\int }_0^1\frac{x}{2x^2-3x-4x+6}dx$$

$$={\int }_0^1\frac{x}{(2x-3)(x-2)}dx$$

$$={\int }_0^1\left[\frac{A}{(2x-3)}+\frac{B}{(x-2)}\right]dx$$

$$={\int }_0^1\left(\frac{-3}{2x-3}+\frac{2}{x-2}\right)dx$$

$$={\left[-\frac{3\log |2x-3|}{2}+2\log |x-2|\right]}_0^1$$

$$=-\frac{3}{2}\left[\log (-1)-\log (-3)\right]+2\left[\log (-1)-\log (-2)\right]$$

$$=-\frac{3}{2}\log \left(\frac{1}{3}\right)+2\log \left(\frac{1}{2}\right)$$

$$=+\frac{3}{2}\log 3-2\log 2$$

$$=\log \sqrt{3^3}-\log 4$$

$$=\log \frac{\sqrt{3^2\times 3}}{4}$$

$$=\log \frac{3\sqrt{3}}{4}$$