Answer
$A=\lim\limits_{n \to \infty}\sum_{i}^n\frac{4x_i^2e^{x_i}}{n}$
Work Step by Step
Using the Definition 2,
$A=\lim\limits_{n \to \infty}\sum_{i=1}^nf(x_i)\Delta x$ where $\Delta x=\frac{upper\ bound\ -\ lower\ bound}{n}$
Find $\Delta x$ and simplify:
$\Delta x=\frac{4-0}{n}=\frac{4}{n}$
Then,
$A=
\lim\limits_{n \to \infty}\sum_{i=1}^nx_i^2e^{x_i}\cdot \frac{4}{n}$
$A=\lim\limits_{n \to \infty}\sum_{i=1}^n\frac{4x_i^2e^{x_i}}{n}$
