Re: Use Riemann sums and a limit to compute the exact area under the curve.

Once I have the form:

$\displaystyle A_n=\frac{(n+1)(11n+1)}{6n^2}$

I know it is of the form:

$\displaystyle A_n=\frac{11n^2+\cdots}{6n^2}$

and since the degree is the same in the numerator and denominator, I know the limit as $\displaystyle n\to\infty$ is simply the ratio of the leading coefficients, so we must have:

$\displaystyle \lim_{n\to\infty}A_n=\frac{11}{6}$

There is nothing wrong with your approach either, I just find it simpler to think of the technique we are taught in precalculus to find horizontal asymptotes.

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Re: Use Riemann sums and a limit to compute the exact area under the curve.

Okay, I took your advice and reworked the problem and voila, now I get the same answer either way! I posted my work in the attached pic file, just for your perusing pleasure (Nerd)

Thanks for all your help, you ROCK! (Rock)

Re: Use Riemann sums and a limit to compute the exact area under the curve.

Quote:

Originally Posted by

**MarkFL2** Once I have the form:

$\displaystyle A_n=\frac{(n+1)(11n+1)}{6n^2}$

I know it is of the form:

$\displaystyle A_n=\frac{11n^2+\cdots}{6n^2}$

and since the degree is the same in the numerator and denominator, I know the limit as $\displaystyle n\to\infty$ is simply the ratio of the leading coefficients, so we must have:

$\displaystyle \lim_{n\to\infty}A_n=\frac{11}{6}$

There is nothing wrong with your approach either, I just find it simpler to think of the technique we are taught in precalculus to find horizontal asymptotes.

Very good advice. It has been eons and eons since I took Algebra, let alone pre-calculus, which explains why I am not very efficient at these yet. I suppose "Hind sight is 20/20" and I should not have dove straight into calculus without some basic refreshers.

Once again, you have saved me much anguish! I truly thank you!! (Clapping)

Re: Use Riemann sums and a limit to compute the exact area under the curve.

Your revised work looks good! Glad to be of assistance. :D