# Calculus – Express This Limit As A Definite Integral. No Interval Given. $\lim\limits_{n\to\infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$

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## calculus – 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 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 $\int_0^1 2(1+2x)\,dx$.

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5.2 The Definite Integral – Calculus Volume 1 | OpenStax – The integral is "the sum expressed with the sigma symbol as "n" is allowed go to infinity". The "sigma" is then changed to the "music note". A "definite" integral is an integral that can be evaluated between two specific values of a variable. "Definate integral" does not mean writing the summation as an integral as n is allowed go to infinity.so we've got a Riemann sum we're going to take the limit as n approaches infinity and the goal of this video is to see if we can rewrite this as a definite integral I encourage you to pause the video and see if you can work through it on your own so let's remind ourselves how a definite integral can relate to a Riemann sum so if I have the definite integral from A to B of f of X f of X DX weFree definite integral calculator – solve definite integrals with all the steps. Type in any integral to get the solution, free steps and graph This website uses cookies to ensure you get the best experience.

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