Homework Help: Forgot how to integrate fractions

1. Nov 3, 2005

Pengwuino

Ok so I have to find the arc length from t=1 to t=2.

$$\begin{array}{l} L = \int_a^b {|r'(t)|dt} \\ |r'(t)| = \frac{{2(1 + 2t^4 )}}{{t^3 }} \\ \end{array}$$

And I have completely forgotten how to integrate fractions.....

Oh wait... i THINK i know what to do. Should i set u=1+2t^4?

Last edited: Nov 3, 2005
2. Nov 3, 2005

Benny

I hope I'm not missing something.

$$\frac{{2\left( {1 + 2t^4 } \right)}}{{t^3 }} = 2\left( {\frac{1}{{t^3 }} + \frac{{2t^4 }}{{t^3 }}} \right)$$ then it should just be the normal power rule right?

3. Nov 3, 2005

dextercioby

No need for any substitution. Benny's right.

Daniel.

4. Nov 3, 2005

Pengwuino

heh, i don't even remember the power rule.

5. Nov 3, 2005

Reshma

Power rule for integration is:
$$\int x^n dx = \frac{x^{n +1}}{n + 1} + C$$

6. Nov 3, 2005

dextercioby

Don't forget the condition

$$\int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}-\{-1\}}$$

Daniel.

7. Nov 3, 2005

Nylex

Also remember that $$\frac{1}{x^n} = x^{-n}$$

8. Mar 6, 2007

disregardthat

I'm sorry to bring this up, but how do you integrate a fraction?

If you have 4x/3, how do you integrate it?

9. Mar 6, 2007

arildno

Remember that a fraction is nothing else than an ordinary number. Also remember that you can write:
$$\frac{4x}{3}=\frac{4}{3}*x$$
f
Thus, an anti-derivative is:
$$\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C$$

10. Mar 6, 2007

Kruger

Also think of the following: If there is a constant inside an integral you can move it outside the integral (sorry, don't know this formula editor):

INTEGRAL(c*f(x)dx)=c*INTEGRAL(f(x)dx)

So in your case 4/3 is a constant (independant of x) and you can move it outside the integral, getting:

INTEGRAL(4x/3dx)=4/3*INTEGRAL(xdx)

11. Mar 7, 2007

disregardthat

Thanks,

$$\int{C*f(x)dx} = C$$ $$\int{f(x)dx}$$

But Kruger, wouldn't the dx be over/beside the fraction, and not under: like this?

$$\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}$$

Last edited: Mar 7, 2007
12. Mar 7, 2007

HallsofIvy

That's what he meant. Most people interpret 4/3a to be (4/3)a, not 4/(3a).
It's always best to use the parentheses, though!

13. Mar 8, 2007

disregardthat

We all love parantheses!

14. Mar 23, 2011

emeraldevan

I want to integrate curve eqn to get volume of curve rotated around x-axis
The curve eqn is 6/(5-2x) and the x limits are 0 and 1

15. Mar 23, 2011

Staff: Mentor

Welcome to Physics Forums!

This thread is more than four years old. When you tack an unrelated question onto an existing thread, that's called "hijacking" the thread. Please use the New Thread button to start a new thread.

16. Mar 23, 2011

Staff: Mentor

Actually, the button is labeled "New Topic" not "New Thread", otherwise Mark's advice is good. Also, please show the work that you have already done on the problem, or at least tell us what is confusing you about how to solve it.