## Calculus 8th Edition

$$f(x)=x^{2}+\sqrt {1+2x}, \quad 4 \leqslant x \leqslant 7$$ 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}
$$f(x)=x^{2}+\sqrt {1+2x}, \quad 4 \leqslant x \leqslant 7$$ 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}