# evaluating a definite integral

• Dec 1st 2006, 08:34 PM
thedoge
evaluating 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, 10:14 PM
Jameson
Ahh I love these questions.

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

Now rewrite to an integral integral with this substitution.

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

Now for the bounds.

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

Got it? :)
• Dec 2nd 2006, 08:08 AM
topsquark
Quote:

Originally Posted by Jameson
Ahh I love these questions.

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

Now rewrite to an integral integral with this substitution.

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

Now for the bounds.

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

Got it? :)

Shouldn't that second bound be $\infty$?

-Dan
• Dec 2nd 2006, 11:18 AM
Jameson
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, 11:36 AM
TD!
For the integral, the limits of x are indeed 0 to 2.
• Dec 2nd 2006, 02:12 PM
thedoge
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, 02:46 PM
ThePerfectHacker
Quote:

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

http://img218.imageshack.us/img218/4903/problemeu5.jpg

The standard form the the Riemann Ingegral is,
$\lim_{n\to \infty} \sum_{k=1}^n f(a+k\Delta x)\Delta x$
In this case,
$\Delta x=\frac{2}{n}$
Thus,
$b-a=2$
But there is not "a" term,
$a=0$ thus, $b=2$.
And the function is,
$f(x)=1+x+x^2$

Thus we need to find,
$\int_0^2 1+x+x^2 dx$
Which we can easily evaluate by Fundamental theorem because this function is continous on the closed interval.
• Dec 2nd 2006, 05:58 PM
topsquark
Quote:

Originally Posted by topsquark
Shouldn't that second bound be $\infty$?

-Dan

:o Momentary brain fart!

-Dan