Riemann sum $ \lim_{n\to \infty}\frac2n\sum_{k=1}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)$ to integral

Question

I don't know how to make the following limit $$\displaystyle \lim_{n\to \infty}\frac2n\sum_{k=1}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)$$

into a definite integral and just need some guiding help anything will help thanks.

Answer 1

We have that

$$\lim_{n\to \infty}\frac1n\sum_{k=0}^{n} f\left(a+{k\over n}(b-a)\right)=\int_a^b f(x) dx$$

and therefore

$$\lim_{n\to \infty}\frac2n\sum_{k=0}^n \left(2+\frac kn\right)\left( \ln\left(2+\frac kn \right)\right)=2\int_0^1(2+x)\ln(2+x)\,dx$$