How would I evaluate this integral using the 2nd Fundamental Theorem of Calculus? Thanks in advance.

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- Dec 1st 2006, 07:34 PMthedogeevaluating a definite integral
How would I evaluate this integral using the 2nd Fundamental Theorem of Calculus? Thanks in advance.

http://img218.imageshack.us/img218/4903/problemeu5.jpg - Dec 1st 2006, 09:14 PMJameson
Ahh I love these questions.

Let $\displaystyle \frac{2i}{n}=x$. Thus $\displaystyle dx=\frac{2}{n}$.

Now rewrite to an integral integral with this substitution.

$\displaystyle \int_{L} 1 +x +x^2dx$

Now for the bounds.

For the first bound let $\displaystyle i=0$ and for the second let $\displaystyle i=n$. So the bounds are from 0 to 2.

Got it? :) - Dec 2nd 2006, 07:08 AMtopsquark
- Dec 2nd 2006, 10:18 AMJameson
I don't believe so. Why would it be inifinity? On these problems i=0, then i=n. Is there another way you do it?

- Dec 2nd 2006, 10:36 AMTD!
For the integral, the limits of x are indeed 0 to 2.

- Dec 2nd 2006, 01:12 PMthedoge
Yea that explanation makes sense Jameson, but I'm still not sure how this would be solved once reaching that point. How'd you determine the bounds were 0 and 2?

- Dec 2nd 2006, 01:46 PMThePerfectHacker

The standard form the the Riemann Ingegral is,

$\displaystyle \lim_{n\to \infty} \sum_{k=1}^n f(a+k\Delta x)\Delta x$

In this case,

$\displaystyle \Delta x=\frac{2}{n}$

Thus,

$\displaystyle b-a=2$

But there is not "a" term,

$\displaystyle a=0$ thus, $\displaystyle b=2$.

And the function is,

$\displaystyle f(x)=1+x+x^2$

Thus we need to find,

$\displaystyle \int_0^2 1+x+x^2 dx$

Which we can easily evaluate by Fundamental theorem because this function is continous on the closed interval. - Dec 2nd 2006, 04:58 PMtopsquark