Forgot how to integrate fractions

[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?

 

[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?

 

[dextercioby]

No need for any substitution. Benny's right.

Daniel.

 

[Pengwuino]

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

 

[Reshma]

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

 

[dextercioby]

Don't forget the condition

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

Daniel.

 

[Nylex]

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

 

[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?

 

[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:
$$\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C$$

 

[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)

 

[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?

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

 

[HallsofIvy]

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?

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

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!

 

[disregardthat]

We all love parantheses!

 

[emeraldevan]

Hi, found this thread and hoping to get a reply:
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

 

[Mark44]

Hi, found this thread and hoping to get a reply:
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

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.

 

[jtbell]

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.