Answer
$$
f(x)=x^{2}+\sqrt {1+2x}, \quad 4 \leqslant x \leqslant 7
$$
According to Definition 2, an expression for the area under the graph of $f$ as a limit is:
$$
\begin{aligned}
A &=\lim _{n \rightarrow \infty} R_{n} \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} (4+3 i / n)^{2}+\sqrt {1+2(4+3 i / n)} \cdot \frac{3}{n}
\end{aligned}
$$
Work Step by Step
$$
f(x)=x^{2}+\sqrt {1+2x}, \quad 4 \leqslant x \leqslant 7
$$
the width of a subinterval is
$$
\Delta x=(7-4) / n=3 / n
$$
and
$$
x_{i}=4+i \Delta x=4+3 i / n
$$
The sum of the areas of the approximating rectangles is
$$
R_{n}=f\left(x_{1}\right) \Delta x+f\left(x_{2}\right) \Delta x+\cdots+f\left(x_{n}\right) \Delta x
$$
According to Definition 2, the area is
$$
\begin{aligned}
A &=\lim _{n \rightarrow \infty} R_{n} \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} (4+3 i / n)^{2}+\sqrt {1+2(4+3 i / n)} \cdot \frac{3}{n}
\end{aligned}
$$
