I am told to evaluate this integral by computing the limit of Riemann sums. From another post (https://math.stackexchange.com/a/890615/502497)
I gathered this statement
The right endpoint Riemann sum for $\displaystyle\int_a^b f(x)\,dx$ is given by $\dfrac{b-a}{n}\displaystyle\sum_{k = 1}^{n}f\left(a+\dfrac{b-a}{n}k\right)$.
Now, figure out what $a,b$ and $f(x)$ need to be to make this sum look like the one in the problem.
Once you do that, if $f(x)$ is continuous, then $\displaystyle\lim_{n \to \infty}\dfrac{b-a}{n}\sum_{k = 1}^{n}f\left(a+\dfrac{b-a}{n}k\right) = \int_a^b f(x)\,dx$.
I have attempted to evaluate the integral by solving the limit of the Reimann sums.
$$\int^{2}_{-2}(x^2-1)dx$$
After applying the formula process above, I result with this.
$$\Delta{x}=\frac{2-(-2)}n=\frac4n$$
$$x_0=-2 \to x_i=-2+\frac{4i}{n}$$
$$\sum_{i = 1}^{n}f(c_i)\Delta{x}=\Delta{x}\sum_{i = 1}^{n}[(-2+\frac{4i}{n})^2-1]=\frac4n\sum_{i = 1}^{n}[(-4+\frac{16i}{n}+\frac{16i^2}{n^2}-1)]$$
$$=\frac4n\sum_{i = 1}^{n}[(-3+\frac{16i}{n}+\frac{16i^2}{n^2})]=\frac4n\sum_{i = 1}^{n}[(3n-\frac{16}{n}+\frac{n(n+1)}{2}+\frac{16}{n^2}+\frac{n(n+1)(2n+1)}{6})]$$
$$=12-\frac{32(n+1)}{n}+\frac{32(2n^2+3n+1)}{n}=\frac43+\frac{32}{n^2}$$
$$\lim_{n \to \infty}(\frac43+\frac{32}{n^2})=\frac43$$
So a few questions:
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What does the $n$ represent in $\frac{32}{n^2}$? Does it cancel out because if $n=0$?
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Is there a specific reason why the right endpoint is used to calculate the limit?
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Can a left endpoint or midpoint be used? Is it the same formula for left? What about the formula for midpoint?
Thanks
Best Answer
$n$ is the number of rectangles you split the area you're trying to find into. The idea here is to make the number of rectangles go to infinity and see what number the limit approaches. So, as $n$ gets larger and larger (in other words, goes to infinity), the expression $\frac{32}{n^2}$ gets closer and closer to zero which in turn makes the expression $\frac{4}{3}+\frac{32}{n^2}$ get closer and closer to $\frac{4}{3}$. And that apparently is supposed to be the numerical value of the area you're looking for.
As far as I know, no. It makes no difference whatsoever which endpoint rule you use to calculate your integral. You will arrive at the same answer regardless.
The formula for the left endpoint rule is the same as that for the right endpoint rule: $a+\frac{b-a}{n}i$. The only difference is that you need to change the index variable in your Riemann sum from 1 to 0: $\sum_{i=0}^{n-1}f(x_{i})\Delta{x}$. And lastly, the formula for the midpoint rule is $a+\left(i-\frac{1}{2}\right)\frac{b-a}{n}$.
Although the way you found the integral is totally fine, I decided to try my hand at calculating it too.
First of all, notice that the function $f(x)=x^2-1$ is even:
$$ f(-x)=(-x)^2-1=x^2-1=f(x) $$
which means that the graph of this function with respect to the y-axis on the right-hand side is geometrically equivalent to that on the left-hand side. So, let's change the bounds of integration from $-2$, $2$ to $0$, $2$ and double the integral: $2\int^{2}_{0}(x^2-1)dx$. Finally, here's my solution:
$$ \Delta{x}=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n} $$
$$ x_{i}=a+\frac{b-a}{n}i=a+\Delta{x}\cdot i=0+\frac{2}{n}i=\frac{2}{n}i $$
\begin{align} 2\int^{2}_{0}\left(x^2-1\right)dx &=2\lim_{n \to \infty}\sum_{i = 1}^{n}f(x_{i})\Delta{x}\\ &=2\lim_{n \to \infty}\sum_{i = 1}^{n}f\left(\frac{2}{n}i\right)\frac{2}{n}\\ &=2\lim_{n \to \infty}\frac{2}{n}\sum_{i=1}^{n}\left[\left(\frac{2}{n}i\right)^2-1\right]\\ &=2\lim_{n \to \infty}\frac{2}{n}\left(\frac{4}{n^2}\sum_{i=1}^{n}i^2-\sum_{i=1}^{n}1\right)\\ &=2\lim_{n \to \infty}\frac{2}{n}\left(\frac{4}{n^2}\cdot\frac{2n^3+3n^2+n}{6}-n\right)\\ &=2\lim_{n \to \infty}\left[\frac{8}{n^3}\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)-2\right]\\ &=2\lim_{n \to \infty}\left(\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}-2\right)\\ &=2\left(\frac{8}{3}+0+0-2\right)\\ &=2\left(\frac{8}{3}-\frac{6}{3}\right)\\ &=\frac{4}{3}\\ \end{align}