Evaluate the integral by computing the limit of Riemann sums.

**JDS** · December 3rd 2012, 07:27 PM

I am hoping someone can critique my work and see if I did this problem correctly. Thanks in advance!

NOTE: I have attached a picture file that shows my work, I hope that's okay)

**MarkFL** · December 3rd 2012, 09:22 PM

Now, you didn't do it correctly...you need to subtract the areas of the rectangles that are below the x-axis.

I would suggest using the symmetry of the function (function is even) to simplify your computations, i.e.:

$\int_{-2}^2 x^2-1\,dx=2\int_0^2 x^2-1\,dx$ (the even-function rule).

Now you have the interval [0,1) where the function is negative and the interval (1,2] where the function is positive.

**JDS** · December 4th 2012, 08:02 AM

Okay thanks for pointing that out, I had forgot that I needed to do that. I reworked the problem and am attaching my work as a pic file. If you can, please let me know if Ive made any mistakes, Thanks!!

**JDS** · December 4th 2012, 04:12 PM

Anyone care to assist me by taking a look at my worked out problem above, ... Thanks in advance

**MarkFL** · December 4th 2012, 06:17 PM

Your result is 10 times what it should be.

I would set it up as follows:

$A_n=2\left(\frac{2}{n}\sum_{k=0}^{n-1}\left(\left( \frac{2k}{n} \right)^2-1 \right) \right)=$

$\frac{4}{n}\left(\frac{4}{n^2}\sum_{k=0}^{n-1}(k^2)-n \right)=$

$\frac{16}{n^3}\cdot\frac{(n-1)(n)(2n-1)}{6}-\frac{4n}{n}=$

$\frac{8(n-1)(2n-1)}{3n^2}-4=$

$\frac{8(n-1)(2n-1)-12n^2}{3n^2}=$

$\frac{4(n^2-6n+2)}{3n^2}$

Hence:

$\lim_{n\to\infty}A_n=\frac{4}{3}$

**JDS** · December 4th 2012, 07:52 PM

wow, not sure how i managed to do that, but thanks for the clarification.

**MarkFL** · December 4th 2012, 08:03 PM

My advice about needing to subtract was erroneous, the summation takes care of all that. Also, the even function rule for integration was unnecessary too, as you see, we wind up with the correct $\Delta x=\frac{4}{n}$ either way.

**MarkFL** · December 4th 2012, 09:26 PM

I wanted to point out that I have now changed the upper limit of summation above to n-1 from n. while this doesn't change the end result, it does now give the correct value for $A_n$ .

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