# Question #d7427

Feb 10, 2017

There are some details in the definition that can be changed by authors, but here are a couple of possible answers.

#### Explanation:

The area under the graph of $f$ and above the $x$ axis $a \le x \le b$

Can be defined as:

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i} \text{*}\right) \Delta x$

Where, for each positive integer $n$, we let $\Delta x = \frac{b - a}{n}$

And for $i = 1 , 2 , 3 , . . . , n$ we let ${x}_{i} \text{*}$ be any number in the ${i}^{\text{th}}$ subinterval. That is, ${x}_{i} \text{*}$ is in $\left[a + \left(i - 1\right) \Delta x , a + i \Delta x\right]$. This interval is often denoted $\left[{x}_{i - 1} , {x}_{i}\right]$.

Alternatively
for $i = 1 , 2 , 3 , . . . , n$, we let ${x}_{i} \text{*} = a + i \Delta x$. (These ${x}_{i} \text{*}$'s are the right endpoints of the subintervals.)

In this problem , we have

$f \left(x\right) = {x}^{2} + \sqrt{1 + 2 x}$, and

$a = 7$ and $b = 9$, so $\Delta x = \frac{2}{n}$.

Using the first definition above we have

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i} \text{*}\right) \Delta x$.

Using the $f$ and $\Delta x$ for this problem we can write

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left(\left({x}_{i} \text{*")^2+sqrt(1+2x_i"*}\right)\right) \frac{2}{n}$.

where each ${x}_{i} \text{*}$ is in $\left[7 + \frac{2 \left(i - 1\right)}{n} , 7 + \frac{2 i}{n}\right]$.

Alternatively

We can use the right endpoints of the subintervals. In which case we get

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$. where for $i = 1 , 2 , 3 , . . . , n$, we let ${x}_{i} = 7 + \frac{2 i}{n}$.

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left(7 + \frac{2 i}{n}\right) \Delta x$

Using the $f$ and $\Delta x$ for this problem we can write

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[\left({\left(7 + \frac{2 i}{n}\right)}^{2} + \sqrt{1 + 2 \left(7 + \frac{2 i}{n}\right)}\right) \frac{2}{n}\right]$

In order to satisfy whoever is evaluating your work, you may need to do some algebra and rewrite this as

$= {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[\left(\left(49 + \frac{28 i}{n} + \frac{4 {i}^{2}}{n} ^ 2 + \sqrt{15 + \frac{4 i}{n}}\right)\right) \frac{2}{n}\right]$

$= {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[\frac{98}{n} + \frac{56 i}{n} ^ 2 + \frac{8 {i}^{2}}{n} ^ 3 + \frac{2}{n} \sqrt{15 + \frac{4 i}{n}}\right]$