# Evaluate the limit for a function defined on [0,1]

The limit is a Riemann sum

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2} \right)$$ $$\delta x=\frac{1}{n}$$, so I distribute it to the terms to get

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^6}+\frac{i}{n^3} \right)$$ Now that they have similar denominators I multiply $$\frac{i}{n^3}\cdot(n^3)$$ to get $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^6}+\frac{2i}{n^6} \right)$$

Combining the terms I get $$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{2i^4}{n^6}\right)$$ Knowing that $$i^2=\frac{n(n+1)(2n+1)}{6}$$ I split up the $$i^4$$ into$$\lim_{n\rightarrow\infty}\frac{2}{n^6}\sum_{i=1}^n\left(\frac{n(n+1)(2n+1)}{6}\right)+ \left(\frac{n(n+1)(2n+1)}{6}\right)$$ Am I on the right track?

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2} \right)=\lim_{n\rightarrow\infty}\frac1n\sum_{i=1}^n\left(\frac{i^4}{n^4}+\frac{i}{n} \right)=\int_0^1(x^4+x) dx$$