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How can I find the area bound by $\;x=0,\, x=1,\;$ the $\;x$-axis ($y = 0$) and $\;y=x^2+2x\;$ using Riemann sums?

I want to use the right-hand sum. Haven't really found any good resources online to explain the estimation of areas bounded by curves, hoping anyone here can help?

By the way, I would like there to be 100 intervals.

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100 intervals? You know you must take the limit of the Riemann sum when $\,n\to\infty\,$ while also the maximal length of subintervals goes to zero, right? So you're definitely going to need way more than 100 intervals... –  DonAntonio Jan 18 '13 at 15:45
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@DonAntonio: I think the asker is referring to Riemann sums to approximate error, as in mathworld.wolfram.com/RiemannSum.html –  amWhy Jan 18 '13 at 15:49
    
Do you mean "right hand sum" to mean using the right endpoint of each $\Delta x$ interval? –  amWhy Jan 18 '13 at 15:50
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2 Answers

up vote 3 down vote accepted

Right Riemann Sums place the right corner of the rectangles on the curve.

Right Riemann Sums are an overestimation of area because of all the extra space that is not under the curve that is still calculated in the area because it is inside the rectangles.

For your problem, we have:

$f(x) = x^{2} + 2x$

With: $a = 0, b = 1$ and $100$ rectangles, we have $\frac{b-a}{n} = \frac{1-0}{100} = \frac{1}{100}$

Our formula now becomes:

$\frac{b-a}{n} [f(0) + f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})]$

$\bullet$ For Left Riemann, plug in all numbers except last.

$\bullet$ For Right Riemann, plug in all numbers except first.

$\bullet$ For Midpoint Riemann, average the numbers and plug those in.

So, using Right Riemann, we have:

$f(\frac{1}{100}) = \frac{1^{2}}{100^{2}} + \frac{2}{100} = \frac{201}{10000}$

$f(\frac{2}{100}) = \frac{2^2}{100^{2}} + \frac{4}{100} = \frac{101}{2500}$

$\cdots$

$f(\frac{100}{100}) = \frac{100^2}{100^{2}} + \frac{200}{100} = 3$

You'll get this set of values using WA.

Now, using the formula above, we have:

$\frac{1}{100}[f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})] = (\frac{1}{100})(\frac{26967}{200}) = 1.34835$

Comparing that with the actual $\int_0^1 (x^{2} + 2x) dx = \frac{4}{3} = 1.33333$, we see, as expected, that we got an overestimate.

You can compare these two with the nice pointer by amWhy at Wolfram Math World.

You should repeat this for left and midpoint and compare so you understand!

Regards

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Thanks for such a detailed answer! –  goddfree Jan 18 '13 at 16:52
    
@CharlesTian: Welcome to MSE and you are very welcome tp our wonderful math community! Regards –  Amzoti Jan 18 '13 at 16:54
    
Nice detail and clarification! –  amWhy May 8 '13 at 2:05
    
@amWhy: Thanks, those are fun problems. –  Amzoti May 8 '13 at 2:31
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For a resource see Mathworld: Riemann Sum.

You can enter the function, the number of intervals ($n = 100; \Delta x = 0.01$), choose "right" from the drop-down menu, and click on "replot" to see your Riemann Sum approximation of area with the curve and intervals graphed.

You'll also find the general formula for the sum: $$\sum_{k = 1}^n f(x_k^*)\Delta x$$

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Thanks for the resource, it's helped a lot :) –  goddfree Jan 18 '13 at 16:53
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