Integration

• Feb 26th 2009, 10:23 AM
HunkaMath
Integration
This is a table...

Concentration of Chemical A, x....|1 | 2| 3 |4 | 5 | 6 |7
Concentration of Chemical B, f(x)|12|16|18|21|24|27|32

Let me know if you need help figuring out that table.

So the 3 questions I am having trouble with are as follows.

a) Plot points and connect them with line segments (I can do this)
b) Use the trapezoidal rule to find the area bounded by the broken line of part a, the x-axis, the line x=1 and line x=7
c)Then approximate the area again using the Simp rule

Thanks!
• Feb 26th 2009, 10:32 AM
running-gag
You have to sum the areas of the 6 trapeziums below
http://nsa05.casimages.com/img/2009/...3439427124.jpg

For the second one that I have indicated in blue :
$\displaystyle A_2 = \frac{(18 + 16) \cdot 1}{2} = 17$
• Feb 26th 2009, 03:17 PM
HunkaMath
Quote:

Originally Posted by running-gag
You have to sum the areas of the 6 trapeziums below
http://nsa05.casimages.com/img/2009/...3439427124.jpg

For the second one that I have indicated in blue :
$\displaystyle A_2 = \frac{(18 + 16) \cdot 1}{2} = 17$

where does the 2 come from in the numerator?
• Feb 26th 2009, 03:29 PM
Reckoner
Quote:

Originally Posted by HunkaMath
where does the 2 come from in the numerator?

There is no 2 in the numerator, but the one in the denominator comes from the formula for the area of a trapezoid (or trapezium, if you are British):

$\displaystyle A = \frac12h(b_1+b_2),$

where $\displaystyle b_1$ and $\displaystyle b_2$ are the lengths of the bases, and $\displaystyle h$ is the height.

Also, rather than find the areas of each trapezoid directly, you can apply the Trapezoidal Rule,

$\displaystyle \int_a^b f(x)\,dx \approx \frac{b-a}{2n} \left(f(x_0) + 2f(x_1) + 2f(x_2)+\cdots+2f(x_{n-1}) + f(x_n) \right),$

as specified in the problem's instructions. For reference, Simpson's Rule is

$\displaystyle \int_a^bf(x)\,dx$

$\displaystyle \approx\frac{b-a}{3n}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x _{n-1})+f(x_n)\right].$

You can easily find derivations of these with a quick search.
• Feb 26th 2009, 03:57 PM
HunkaMath
Quote:

Originally Posted by Reckoner
There is no 2 in the numerator, but the one in the denominator comes from the formula for the area of a trapezoid (or trapezium, if you are British):

$\displaystyle A = \frac12h(b_1+b_2),$

where $\displaystyle b_1$ and $\displaystyle b_2$ are the lengths of the bases, and $\displaystyle h$ is the height.

Also, rather than find the areas of each trapezoid directly, you can apply the Trapezoidal Rule,

$\displaystyle \int_a^b f(x)\,dx \approx \frac{b-a}{2n} \left(f(x_0) + 2f(x_1) + 2f(x_2)+\cdots+2f(x_{n-1}) + f(x_n) \right),$

as specified in the problem's instructions. For reference, Simpson's Rule is

$\displaystyle \int_a^bf(x)\,dx$

$\displaystyle \approx\frac{b-a}{3n}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\cdots+4f(x _{n-1})+f(x_n)\right].$

You can easily find derivations of these with a quick search.

do you know what the answers should be I am getting 132 and I am sure that is wrong. thanks
• Feb 26th 2009, 04:49 PM
Reckoner
Quote:

Originally Posted by HunkaMath
do you know what the answers should be I am getting 132 and I am sure that is wrong. thanks

You should be getting 128. Can you show your work?