Integration

**HunkaMath** · February 26th 2009, 10:23 AM

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!

**running-gag** · February 26th 2009, 10:32 AM

You have to sum the areas of the 6 trapeziums below

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

**HunkaMath** · February 26th 2009, 03:17 PM

Originally Posted by running-gag

You have to sum the areas of the 6 trapeziums below

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

where does the 2 come from in the numerator?

**Reckoner** · February 26th 2009, 03:29 PM

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):

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

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

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

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

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

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

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

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

**HunkaMath** · February 26th 2009, 03:57 PM

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):

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

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

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

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

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

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

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

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

**Reckoner** · February 26th 2009, 04:49 PM

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?

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