Mathematics - Definite Integration Question with Solution | TestHub

The given limit is a Riemann sum.

Let $x_k=k/n$. The general term is $\frac{1}{n}\frac{1}{(k/n+a)(k/n+b)}$.

The sum can be written as ${\int }_0^1\frac{1}{(x+a)(x+b)}dx$.

If $a\ne b$, use partial fractions: $\frac{1}{b-a}{\int }_0^1\left(\frac{1}{x+a}-\frac{1}{x+b}\right)dx=\frac{1}{b-a}[\ln |x+a|-\ln |x+b|{]}_0^1=\frac{1}{b-a}\ln {\left|\frac{x+a}{x+b}\right|}_0^1=\frac{1}{b-a}\left(\ln \frac{1+a}{1+b}-\ln \frac{a}{b}\right)=\frac{1}{b-a}\ln \left(\frac{b(1+a)}{a(1+b)}\right)$.

This matches option B.

If $a=b$, the integral is ${\int }_0^1\frac{1}{(x+a{)}^2}dx={\left[-\frac{1}{x+a}\right]}_0^1=-\frac{1}{1+a}-\left(-\frac{1}{a}\right)=\frac{1}{a}-\frac{1}{1+a}=\frac{1+a-a}{a(1+a)}=\frac{1}{a(1+a)}$.

This matches option D.