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I looked at all the resources for Riemann Sums for BC calculus and I could not find any that solved them like my teacher does. The question asks:

Express the following Riemann Sums as definite integrals

$\lim_{n \to \infty}\sum_{k=1}^{2n}(\frac{1}{1+\frac{2k}{n}}) (\frac{1}{n})$

use the following x- values and define an appropriate definite integral

a. x = $\frac{k}{n}$

b. x = $\frac{2k}{n}$

c. x = $1 +\frac{2k}{n}$

My teacher said that there are five steps to converting a Riemann sum into a definite integral,

  1. Determine a possible dx
  2. Determine a value for x
  3. Determine the bounds of the definite integral
  4. Verify your dx is correct
  5. Determine your function & write the definite integral

So for problem a I said dx = $\frac{1}{n}$

x = $\frac{k}{n}$

Then to determine the lower bound I said the lower limit = $\lim_{x \to \infty} \frac{1}{n}$ = 0

Then to determine the upper bound I said the upper limit = $\lim_{x \to \infty} \frac{2n}{n}$ = 2

To verify dx I said dx = $\frac{b - a}{n}$ = $\frac{2 - 0}{2n}$ = $\frac{1}{n}$

Then the function is $\frac{1}{1+2x}$, so my definate integral is $\int_{0}^2 \frac{1}{1+2x}dx$

I have no idea how to do b or c or even if my answer to a is correct. All help is appreciated thanks in advance.

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  • $\begingroup$ Do you mean $\lim_{n \to \infty}\sum_{k=1}^{n}(\frac{1}{1+\frac{2k}{n}}) (\frac{1}{n})$, where I put $\sum_{k=1}^{n}$ not $\sum_{k=1}^{2n}$? $\endgroup$ – user236182 Nov 4 '17 at 18:08
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Your answer to part a is correct.

For part b: $$ dx= x_{k+1}-x_k$$ $$=\frac{2(k+1)}{n}-\frac{2k}{n}=\frac{2}{n}$$

About the limits, ($a$=lower limit, $b$=upper limit) $$a=\lim_{n\to\infty}\frac{2(1)}{n}=0,$$

$$b=\lim_{n\to \infty}\frac{2(2n)}{n}=4$$ Therefore , the integral would be: $$I=\int_0^4\frac{1}{1+x}(0.5dx)$$or $$I=\int_0^4\frac{0.5}{1+x}dx$$

Same drill for part c(will skip some steps): $$ dx=x_{k+1}-x_k=(1+\frac{2(k+1)}{n})-(1+\frac{2k}{n})=\frac{2}{n}$$ $$a=\lim_{n\to \infty}1+\frac{2(1)}{n}=1$$ $$b=\lim_{n\to \infty}1+\frac{2(2n)}{n}=5$$ $$I=\int_1^5\frac{1}{x}(0.5dx)=\int_1^5\frac{0.5}{x}dx $$

Notice that irrespective of the choice of $x,dx$ and limits ( $a$ and $b$), the definite integral evaluates to the same number.

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