Mathematics - Definite Integration Question with Solution | TestHub

The given limit can be written as:$$\underset{n\to \infty }{\lim }\sum _{k=n+1}^{2n}\frac{k^2}{n^3+k^3}=\underset{n\to \infty }{\lim }\sum _{k=n+1}^{2n}\frac{1}{n}\frac{(k/n{)}^2}{1+(k/n{)}^3}$$ This is a Riemann sum. Let $x=k/n$ and $dx=1/n$. As $n\to \infty$, the sum becomes a definite integral. The lower limit for $x$ is ${\lim }_{n\to \infty }(n+1)/n=1$. The upper limit for $x$ is ${\lim }_{n\to \infty }2n/n=2$. So the limit is equal to the integral:$${\int }_1^2\frac{x^2}{1+x^3}dx$$ To evaluate this integral, let $u=1+x^3$. Then $du=3x^{2}dx$, so $x^{2}dx=\frac{1}{3}du$. When $x=1$, $u=1+1^3=2$. When $x=2$, $u=1+2^3=1+8=9$. Substituting these into the integral:$${\int }_2^9\frac{1}{u}\frac{1}{3}du=\frac{1}{3}[\ln |u|{]}_2^9=\frac{1}{3}(\ln 9-\ln 2)=\frac{1}{3}\ln \left(\frac{9}{2}\right)$$ Comparing this with $\frac{1}{P}\ln \left(\frac{Q}{R}\right)$, we have $P=3,Q=9,R=2$. Thus, $P+Q+R=3+9+2=14$.