Answer
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}\left(\left(i\frac{4}{n}\right)^{2}+3\right)\frac{4}{n}$
where:
$\Delta x=\frac{4}{n}$
$x_{i}=i\frac{4}{n}$
Work Step by Step
$$\int_{0}^{4}(x^{2}+3)dx$$
So $a=0$, $b=4$ and $f(x)=x^{2}+3$
so:
$$\Delta x=\frac{b-a}{n}=\frac{4-0}{n}=\frac{4}{n}$$
$$x_{i}=a+i\Delta x=0+i\frac{4}{n}=i\frac{4}{n}$$
so:
$$\int_{0}^{4}(x^{2}+3)dx=\lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_{i})\Delta x=\lim\limits_{n \to \infty}\sum_{i=1}^{n}(x_{i}^{2}+3)\Delta x=\lim\limits_{n \to \infty}\sum_{i=1}^{n}\left(\left(i\frac{4}{n}\right)^{2}+3\right)\frac{4}{n}$$
