Reimann's sums

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December 9th 2007, 12:20 PM
akhayoon

Reimann's sums

$\int_{1}^{3}x^2-x+1~dx$

I have to find this integral using reimann's sums

usually what I do si subtract 3-1/n to get 2/n

and that gives me 2i/n which I then sub into the polynomial and multiply by 2/n to give me the answer(using limits and sums ofcourse)...but for this one it's not working so I'm guessing I shouldn't be using 2i/n???:confused:
December 9th 2007, 12:34 PM
Jhevon

Quote:

Originally Posted by akhayoon

$\int_{1}^{3}x^2-x+1~dx$

I have to find this integral using reimann's sums

usually what I do si subtract 3-1/n to get 2/n

and that gives me 2i/n which I then sub into the polynomial and multiply by 2/n to give me the answer(using limits and sums ofcourse)...but for this one it's not working so I'm guessing I shouldn't be using 2i/n???:confused:

no, it is not 2i/n

you have: $x_0 = 1, x_1 = 1 + \frac 2n, x_2 = 1 + \frac 4n \cdots x_i = 1 + \frac {2i}n$

and of course, $\int_a^b f(x)~dx = \lim_{n \to \infty} \sum_{i = 0}^{n} f(x_i) \Delta x$ , where $\Delta x = \frac {b - a}n$
December 9th 2007, 12:41 PM
akhayoon

thanks but I don't understand how you got

$x_1=1+\frac{2}{n}$ or

$x_2=1+\frac{4}{n}$

could you please show me how you got these?(Doh)
December 9th 2007, 12:47 PM
Jhevon

Quote:

Originally Posted by akhayoon

thanks but I don't understand how you got

$x_1=1+\frac{2}{n}$ or

$x_2=1+\frac{4}{n}$

could you please show me how you got these?(Doh)

remember, the $x_i$ 's are the points on the x-axis that are $\Delta x$ apart.

so if $x_0 = a$

then $x_1 = a + \Delta x$ .....since we move a distance of $\Delta x$ to the right to go from $x_0$ to $x_1$

and $x_2 = x_1 + \Delta x$

and $x_3 = x_2 + \Delta x$

and so on.

here, $\Delta x = \frac 2n$

so, $x_0 = 1$

and $x_1 = 1 + \frac 2n$

and $x_2 = x_1 + \Delta x = 1 + \frac 2n + \frac 2n = 1 + \frac 4n$

and $x_3 = x_2 + \Delta x = 1 + \frac 4n + \frac 2n = 1 + \frac 6n$

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$x_i = 1 + \frac {2i}n$