Question

Integral question when you have something minus something else i.e., integral 48-12x^2.

Answer 1

Should you always break it up into \[\int\limits - \int\limits\]

\[\int\limits 48 dx - \int\limits 12x^2 dx\]

Answer 2

Usually makes it easier.

Answer 3

no you don't have to, people do this so it becomes easier for them to see what to do, but it's not a law you have to apply.

Answer 4

Can't think of a counterexample, though.

Answer 5

Okay, because the question was find the average of

that from -5<x<5

Answer 6

so it became \[\frac{1}{10} \int\limits 48-12x^2dx\]

Answer 7

I figured I could take a 12 out so it would become \[\frac{6}{5} \int\limits 4-x^2dx\]

Answer 8

u = 4-x^2dx
1/2 du = xdx

Answer 9

why would you use a substitution here?

Answer 10

Hold on a second

Answer 11

Actually, I guess I could just split them up again huh.

Answer 12

\[\int\limits 4 - \int\limits x^2\]

Answer 13

also don't forget the domains of your integral, you want to evaluate it.

Answer 14

-5 to 5

Answer 15

Ic, so that's where I made my mistake I guess.

Answer 16

so 4x-x^3/3 from -5 to 5

Answer 17

So @KonradZuse slow down a second OK?

Answer 18

sure.

Answer 19

Sorry I'm slow with the equation editor.

Answer 20

You have the integral of 48 - 12x^2 between limits -5 to 5, right?

Answer 21

Why not leave the constants in for now. The best thing to do is to take the two terms separately as you did.
integral of 48 dx
- integral of 12 x^2 dx
right?

Answer 22

yup.. I know what the answer to that is, I was just trying to figure out if we got the same answer goign the other way is all...

Answer 23

What would be the other way? How would you integrate (48 + 12x^2) dx?
You don't have x

Answer 24

Sorry, maybe I just got in your way. Anyhow, I haven't thought of a situation where you wouldn't split em (unless somehow factoring gave you something you needed for a substitution ..

Answer 25

maybe I am now confused myself, it's getting late here, but integrating (48+12x^2)dx directly is possible

Answer 26

Um.. let u = 48 + 12^2
You don't have the x for du

Answer 27

well I understand that, so maybe we do that here a bit different then, but we let such integrals stay the way they are, keeping mind that the entire term in the brackets has to be integrated with respect to dx and then you just get the regular solution.

Answer 28

I agree that it's not mathematically rigorous, but safes at least some writing for small terms.

Answer 29

Maybe I don't understand your example. Can you show me?

Answer 30

\[ \int (48 +12x^3)dx\]

Brackets just say that everything inside is multiplied by dx, but instead of writing that out you can just do it directly, it's the same thing as you do, just wont require splitting it up, but as I said, it loses some rigor definitions about the integral and its function as a linear operator.

\[ \int 48dx+12x^3dx= 48x+3x^4\]

Answer 31

Well, OK. You are splitting it up, just saving yourself actually writing the second integral sign. I can live with that. Personally I find the parentheses confusing because then I'm thinking that is u(x) and I need du..

Answer 32

Yes I understand that, I see the above method applied all the time, without even carrying out the dx (second step I did), but yes it is mainly for saving some time for 'easier' integrals like this, linear exponent terms.

I just asked because it's interesting for me how people apply various rules over the globe (-: there are some little differences.

Answer 33

Great chatting. Get some sleep.

Answer 34

Thanks goodniught.