# approximate the area using Riemann sums

• Jan 11th 2009, 01:04 PM
omgeezy
approximate the area using Riemann sums
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 $\displaystyle (1/3*x^2)$ and the 2nd be 27 by plugging in 9?

Each rectangle's... width would be 2...
something like that?
• Jan 11th 2009, 01:22 PM
Jester
Quote:

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 $\displaystyle \Delta x$ is

$\displaystyle \Delta x = 8/4 = 2$

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

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

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

then the area is approximately

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

Then simplify.
• Jan 11th 2009, 01:28 PM
omgeezy
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
• Jan 11th 2009, 01:52 PM
omgeezy
in the end I have

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

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

right or wrong?
just checking for any mathematical errors
• Jan 11th 2009, 02:15 PM
skeeter
left sum = 72
• Jan 11th 2009, 02:22 PM
omgeezy
Quote:

Originally Posted by skeeter
left sum = 72

What did I miss?
• Jan 11th 2009, 02:37 PM
skeeter
you didn't calculate it correctly. try again.

$\displaystyle 2\left(\frac{1^2}{3} + 2 + \frac{3^2}{3} + 2 + \frac{5^2}{3} + 2 + \frac{7^2}{3} + 2 \right) = 72$
• Jan 11th 2009, 02:52 PM
omgeezy
I see, I missed the $\displaystyle ( \frac{1}{3}\,1^2+2)$ what a stupid mistake
now that makes sense
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