# Math Help - integration :quick question

1. ## integration :quick question

what do I get when I integrate
x^2.5 dx

2. Originally Posted by wolfhound
what do I get when I integrate
x^2.5 dx

Integration without bounds is equivalent to solving for an antiderivative.
Antiderivative of functions of the form, $f(x)=x^n \;where\; n\; is\; real$, amounts to

$Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C$

3. Originally Posted by vince
Integration without bounds is equivalent to solving for an antiderivative.
Antiderivative of functions of the form, $f(x)=x^n$ where n is real, amounts to

$Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C$
Sorry but I don't have a clue what your on about!!!!!!!!
Could someone please show me the answer this part is delaying me

4. Originally Posted by wolfhound
Sorry but I don't have a clue what your on about!!!!!!!!
Could someone please show me the answer this part is delaying me

dude, $\int x^{2.5}dx = \frac{x^{2.5+1}}{2.5+1}+C$...clear now?

5. My problem is what do I raise x^2.5 to? (x^?/?)

6. so its x to the power of 3.5 divided by 3.5 + C ?

7. $\frac{x^{3.5}}{3.5}$ + C

8. Originally Posted by vince
Integration without bounds is equivalent to solving for an antiderivative.
Antiderivative of functions of the form, $f(x)=x^n \;where\; n\; is\; real$, amounts to

$Antiderivative \;of\;x^n = \int x^ndx = \frac{x^{n+1}}{n+1}+C$
$n \neq -1$.

9. Originally Posted by Ted
$n \neq -1$.

but of course, making sure if anyone was paying attention.

10. But you said where n is real.