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,
- Determine a possible dx
- Determine a value for x
- Determine the bounds of the definite integral
- Verify your dx is correct
- 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.
Answer
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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