Calculate the following:
The limit as n goes to infinity, n * integral (from 1 to n) of xdx / (n^(3) + [x]^(2) )

Question

Calculate the following:
The limit as n goes to infinity, n * integral (from 1 to n) of  xdx / (n^(3) + [x]^(2) )

anonymous · Answer

$$
g(x)=)\int \frac{x}{n^3+x^2} \, dx=\frac{1}{2} \ln \left(n^3+x^2\right)\
g(n)-g(1)=\frac{1}{2} \ln \left(n^3+n^2\right)-\frac{1}{2} \ln \left(n^3+1\right)=\
\frac{1}{2} \ln \left(\frac{n^2 (n+1)}{(n+1) \left(n^2-n+1\right)}\right)=\
\frac{1}{2} \ln \left(\frac{n^2 }{ n^2-n+1 }\right)\
\text { Now}\

 \lim_{n->\infty}\frac{1}{2}n \ln \left(\frac{n^2 }{ n^2-n+1 }\right)=\frac 1 2

$$
For the last limit use L'Hospital's Rule