approximate the area using Riemann sums

**omgeezy** · January 11th 2009, 01:04 PM

I need to approximate int((1/3*x^2+2) dx) on the interval [1, 9] using a left Riemann sum, and 4 rectangles of equal base.

I'm having trouble graphing it, I've used my calculator to see what it looks like but if I try to graph it by hand it won't be as perfect and the rectangles wouldn't fit under the curve. The graphing part is really stressing me out.

Would the 1st endpoint be at 2/3? by plugging in 1 into $(1/3*x^2)$ and the 2nd be 27 by plugging in 9?

Each rectangle's... width would be 2...
something like that?

**Danny** · January 11th 2009, 01:22 PM

Originally Posted by omgeezy

I need to approximate int((1/3*x^2+2) dx) on the interval [1, 9] using a left Riemann sum, and 4 rectangles of equal base.

I'm having trouble graphing it, I've used my calculator to see what it looks like but if I try to graph it by hand it won't be as perfect and the rectangles wouldn't fit under the curve. The graphing part is really stressing me out.

The inteval has length 8 so the width of each rectangle $\Delta x$ is

$\Delta x = 8/4 = 2$

When we split up the interval $[1,9]$ into 4 and use the left endpoint, we obtain

$x_1 = 1,\;\;x_2 = 3,\;\;x_3=5,\;\;x_4 = 9$

If we let $f(x) = \frac{1}{3}\,x^2+2$

then the area is approximately

$\sum_{i=1}^4 f(x_i) \Delta x = \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right) \Delta x$
$= \left( f(1) + f(3) + f(5) + f(7) \right) 2$
$= \left( \frac{1}{3}\,1^2+2+ \frac{1}{3}\,3^2+2 + \frac{1}{3}\,5^2+2 + \frac{1}{3}\,7^2+2 \right) 2$

Then simplify.

**omgeezy** · January 11th 2009, 01:28 PM

If it isn't too much trouble, what would the graph look like?

OR is there no need for a graph? maybe that's why

thanks very much

**omgeezy** · January 11th 2009, 01:52 PM

in the end I have

$2(\frac{2}{3}+5+\frac{31}{3}+\frac{55}{3})=\frac{2 06}{3}$

$\frac{206}{3}=68.6667$

right or wrong?
just checking for any mathematical errors

**skeeter** · January 11th 2009, 02:15 PM

left sum = 72

**omgeezy** · January 11th 2009, 02:22 PM

Originally Posted by skeeter

left sum = 72

What did I miss?

**skeeter** · January 11th 2009, 02:37 PM

you didn't calculate it correctly. try again.

$2\left(\frac{1^2}{3} + 2 + \frac{3^2}{3} + 2 + \frac{5^2}{3} + 2 + \frac{7^2}{3} + 2 \right) = 72$

**omgeezy** · January 11th 2009, 02:52 PM

I see, I missed the $( \frac{1}{3}\,1^2+2)$ what a stupid mistake
now that makes sense

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