Express this limit as a definite integral. No interval given. $\lim\limits_{n\to\infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$ I am having trouble trying to convert a limit to a definite integral. I am unsure about how to go about this. I have already tried googling this but can not find anything that is comprehensive enough for me to learn from.
Here's the limit: 
$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$$
I need to express this as a definite integral but cannot figure out how. My textbook is not clear and doesn't include an example, and my professor did not explain this.
Thank you!
 A: The goal is to represent the limit
$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$$
as an integral.
In fact, any integral like $\int_a^b f(x) dx$ can be approximated as a sum of $n$ rectangles:
$$\int_a^b f(x)dx \approx \sum_{k=1}^n f(a + k\cdot\Delta x)\cdot \Delta x$$
A picture shows why— here, $\Delta x$ is the width of the rectangles (it's equal to the length of the interval divided into $n$ equal pieces), $(a+k\Delta x)$ is the x-position of the $k$th rectangle, and $f(a+k\cdot \Delta x)$ is its height so that the left tip of the rectangle touches the curve $f(x)$.
If we increase the number of rectangles $n$, the sum should become a more and more accurate approximation of the integral. Eventually, if the limit exists, the approximation will become exact:
$$\int_a^b f(x)dx = \lim_{n\rightarrow \infty}\sum_{k=1}^n f(a + k\cdot\Delta x)\cdot \Delta x$$
If we match this general pattern against the equation you're given, it looks like:


*

*$\Delta x \longleftrightarrow \frac{2}{n}$ is the rectangle width.

*$f(a + k \Delta x) \longleftrightarrow (1 + k\cdot \frac{2}{n})$

*So the left endpoint $a$ is 1. 

*And $f(x) = x$, nothing more complicated.

*And we can solve for the right endpoint $b$ because we know that $\Delta x \equiv \frac{b-a}{n}$ by definition of these equally-spaced rectangles and $a = 1$ as we have found. So $\frac{2}{n} = \frac{b-1}{n}$ so $b=3$.


We now have all of our components and can write
$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+ \frac{2k}{n}\right)\cdot \frac{2}{n} = \fbox{$\int_{1}^3 x \, dx$}$$
A: If the Riemann integral $\int_0^1 f(x)\,dx$ exists, then it can be written as the limit
$$\int_a^b f(x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n f\left(a+\frac{b-a}{n}\,k\right)\,\left(\frac{b-a}{n}\right)\tag 1$$
Using $(1)$ with $f(x)=2(1+2x)$, $a=0$ and $b=1$ reveals that
$$\int_0^1 2(1+2x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n 2\left(1+\frac {2k}n\right)\,\frac1n=\lim_{n\to \infty}\sum_{k=1}^n \left(1+\frac {2k}n\right)\,\frac2n$$
Therefore, the limit of interest is simply the Riemann Sum of the integral $2\int_0^1 (1+2x)\,dx$.
A: I would like to correct the answer above/below me, with my own comments.  
If the Riemann integral $\int_a^b f(x)\,dx$ exists, then it can be written as the limit of a special sum known as a Riemann sum 
$$\int_a^b f(x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n f(c_k) \Delta x  \tag 1 $$
where  $c_k = a + \frac{b-a}{n} \cdot k $ and $ \Delta x = \frac{b-a}{n}$. The formula for $c_k$ are right endpoints of each of the n uniform width subintervals.
Note that the choice of $c_k$ is not unique and different $c_k$ will produce different functions with different limits for the integral. However the final value for the definite integral should end up being the same.
I will choose $c_k= 0 + \frac{1-0}{n} \cdot k = \frac{k}{n} $ which forces $ \Delta x = \frac{1-0}{n} = \frac 1 n $. 
It may seem more natural to choose $c_k= 1 + \frac{3-1}{n} ~ k$ and $ \Delta x = \frac 2 n $ . This will lead to the first answer posted above. I will leave it to you to read that answer.
Using some algebra we can rewrite the original Riemann sum in the appropriate 'integral ready' form using our choice $c_k= \frac{k}{n}$ and $ \Delta x =\frac 1 n $:  
$$\begin{align}\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n} &= \lim_{n\rightarrow \infty}\sum_{k=1}^n 2\left(1+ 2 \cdot \frac{k}{n}\right)\cdot \frac{1}{n}
\\ &=\lim_{n\rightarrow \infty}\sum_{k=1}^n 2\left(1+ 2 \cdot c_k\right)\cdot \Delta x  
\\ &= \lim_{n\rightarrow \infty}\sum_{k=1}^n f(c_k) \cdot \Delta x
\\ &= \int_{0}^{1} f(x) ~dx  \end{align} $$
Notice how the $c_k$ becomes the $x$ in the definite integral.
It follows the Riemann Sum is equal to the integral $\int_0^1 2(1+2x)\,dx$.
