# Thread: Midpoint Approx. of Integral

1. ## Midpoint Approx. of Integral

I'm not exactly sure how to implement the formula for this.

$\displaystyle \int{x^2 + 1} dx$ from 0 to 4

And the question wants MID(2), which I assume means to divide it into two subsections.

How would I find this?

So far I have drawn the graph and gotten delta x to be 1. So I have

$\displaystyle f(2 + \frac{2}{2})(1) + f(10 + \frac{2}{2})(1)$
$\displaystyle = f(5/2) + f(21/2)$
$\displaystyle = (\frac{5}{2})^2 + 1) + ((\frac{21}{2})^2 + 1)$

according to the equation given to me in my notes, which was:

$\displaystyle \int{f(x)} = f(x_1 + \frac{delta x_1} {2}) (delta x_1) + f (x_2 + \frac{delta x_2} {2})(delta x_2)$

So I got 118.5. Is that right?

2. Originally Posted by lysserloo
I'm not exactly sure how to implement the formula for this.

$\displaystyle \int{x^2 + 1} dx$ from 0 to 4

And the question wants MID(2), which I assume means to divide it into two subsections.

How would I find this?

So far I have drawn the graph and gotten delta x to be 1. So I have

$\displaystyle f(2 + \frac{2}{2})(1) + f(10 + \frac{2}{2})(1)$
$\displaystyle = f(5/2) + f(21/2)$
$\displaystyle = (\frac{5}{2})^2 + 1) + ((\frac{21}{2})^2 + 1)$

according to the equation given to me in my notes, which was:

$\displaystyle \int{f(x)} = f(x_1 + \frac{delta x_1} {2}) (delta x_1) + f (x_2 + \frac{delta x_2} {2})(delta x_2)$

So I got 118.5. Is that right?
This may sound strange, but could you define MID2? I'm not quite certain what you mean?

3. MID(2) means n = 2 I believe. Divided into 2 subsections, so like from 0 to 2, then 2 to 4.

4. Originally Posted by lysserloo
MID(2) means n = 2 I believe. Divided into 2 subsections, so like from 0 to 2, then 2 to 4.
So like $\displaystyle \frac{1}{b-a}\int_a^b{f(x)}dx$?

5. No it's the equation I gave above, using the midpoint to estimate the integral.

Well, actually yours may be the same. I'm really not sure.

Either way, I'm looking for a midpoint approximation, with delta x being
$\displaystyle \frac{b -a}{n}$ and n = 2 I think.

6. Originally Posted by lysserloo
No it's the equation I gave above, using the midpoint to estimate the integral.

Well, actually yours may be the same. I'm really not sure.

Either way, I'm looking for a midpoint approximation, with delta x being
$\displaystyle \frac{b -a}{n}$ and n = 2 I think.
Oh, I see...So, you want

$\displaystyle \int_0^4f(x)dx=\int_0^4(x^2+1)dx\approx\sum_{i=1}^ 2\left[f\left(x_i+\frac{\Delta{x}}{2}\right)\right]{\Delta{x}}$

where $\displaystyle x_1=0,x_2=2$ and $\displaystyle \frac{\Delta{x}}{2}=\frac{4-0}{2}=2$?

7. Yeah, that looks right.

8. Originally Posted by lysserloo
Yeah, that looks right.
Well then

$\displaystyle \int_0^4f(x)dx\approx{f(2)\Delta{x}+f(4)\Delta{x}}$