Forgot how to integrate fractions


by Pengwuino
Tags: forgot, fractions, integrate
Pengwuino
Pengwuino is offline
#1
Nov3-05, 04:03 AM
PF Gold
Pengwuino's Avatar
P: 7,125
Ok so I have to find the arc length from t=1 to t=2.

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

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?
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
Benny
Benny is offline
#2
Nov3-05, 04:07 AM
P: 585
I hope I'm not missing something.

[tex]\frac{{2\left( {1 + 2t^4 } \right)}}{{t^3 }} = 2\left( {\frac{1}{{t^3 }} + \frac{{2t^4 }}{{t^3 }}} \right)[/tex] then it should just be the normal power rule right?
dextercioby
dextercioby is offline
#3
Nov3-05, 04:16 AM
Sci Advisor
HW Helper
P: 11,863
No need for any substitution. Benny's right.

Daniel.

Pengwuino
Pengwuino is offline
#4
Nov3-05, 04:20 AM
PF Gold
Pengwuino's Avatar
P: 7,125

Forgot how to integrate fractions


heh, i don't even remember the power rule.
Reshma
Reshma is offline
#5
Nov3-05, 04:33 AM
P: 777
Power rule for integration is:
[tex]\int x^n dx = \frac{x^{n +1}}{n + 1} + C[/tex]
dextercioby
dextercioby is offline
#6
Nov3-05, 04:43 AM
Sci Advisor
HW Helper
P: 11,863
Don't forget the condition

[tex] \int x^{n} \ dx=\frac{x^{n+1}}{n+1} +C , n\in{\mathbb{C}-\{-1\}} [/tex]

Daniel.
Nylex
Nylex is offline
#7
Nov3-05, 05:31 AM
P: 554
Also remember that [tex]\frac{1}{x^n} = x^{-n}[/tex]
disregardthat
disregardthat is offline
#8
Mar6-07, 03:35 PM
Sci Advisor
P: 1,686
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
arildno is offline
#9
Mar6-07, 03:40 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Remember that a fraction is nothing else than an ordinary number. Also remember that you can write:
[tex]\frac{4x}{3}=\frac{4}{3}*x[/tex]
f
Thus, an anti-derivative is:
[tex]\frac{4}{3}*\frac{x^{2}}{2}+C=\frac{4}{3*2}x^{2}+C=\frac{2}{3}x^{2}+C[/tex]
Kruger
Kruger is offline
#10
Mar6-07, 05:00 PM
P: 219
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
disregardthat is offline
#11
Mar7-07, 11:00 AM
Sci Advisor
P: 1,686
Thanks,

[tex]\int{C*f(x)dx} = C[/tex] [tex]\int{f(x)dx}[/tex]

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

[tex]\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}[/tex]
HallsofIvy
HallsofIvy is offline
#12
Mar7-07, 01:35 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
Quote Quote by Jarle View Post
Thanks,

[tex]\int{C*f(x)dx} = C[/tex] [tex]\int{f(x)dx}[/tex]

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

[tex]\int{\frac{4x}{3}dx} = \frac{4}{3}\int{x*dx}[/tex]

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
disregardthat is offline
#13
Mar8-07, 08:40 AM
Sci Advisor
P: 1,686
We all love parantheses!
emeraldevan
emeraldevan is offline
#14
Mar23-11, 05:05 PM
P: 1
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
Mark44 is online now
#15
Mar23-11, 05:14 PM
Mentor
P: 20,968
Quote Quote by emeraldevan View Post
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
jtbell is offline
#16
Mar23-11, 09:37 PM
Mentor
jtbell's Avatar
P: 11,227
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.


Register to reply

Related Discussions
I forgot How to do these.... Precalculus Mathematics Homework 12
Integrate using Partial Fractions Calculus & Beyond Homework 2
forgot how to integrate!! yes! t*cos(Pi*t) Calculus & Beyond Homework 6
I forgot that I never forget General Discussion 11
Really Simple... I just forgot Introductory Physics Homework 9