Is there a math boffin out there who can help me...? TIA

[Dougster]

Just one quick question:

If, to find the differential of (ax + b)^n it is na(ax + b)^n-1

How do you find the integral of (ax + b)^n?

It's very much appreciated if you can help... my text book has the most long-winded explanation you have ever seen, and I KNOW I used to be able to do it a simple way.

 

[br0wn]

The derivative of (ax + b)^n with respect to x is
n*(ax)^(n-1). It is not n*(ax+b)^(n-1)

The integral of (ax + b) with respect to x is
1/2 *(ax^2) + bx + c, where c is a constant

 

[JayHu]

<< The derivative of (ax + b)^n with respect to x is
n*(ax)^(n-1). It is not n*(ax+b)^(n-1)

The integral of (ax + b) with respect to x is
1/2 *(ax^2) + bx + c, where c is a constant >>



I dont agree with what you said first.
(ax+b)^n is a chain rule
derivative of x^n = nx^(n-1)
so
(ax+b)^n/dx = an( ax+b)^(n-1)

 

[Aelus]

D((ax + b)^n)= na(ax+b)^(n-1)
int((ax + b)dx)=1/2ax^2+bx

my math professor would kill me if i have those wrong.

Aelus

 

[Dougster]

Derivative! That's the word I was looking for, I knew "differential" didn't sound right!

JayHu, you are correct, this does come from the chain rule, and I know that D((ax + b)^n)= na(ax+b)^(n-1), as Aelus put it, I copied it straight out of a text book.

Aelus, thanks for the answer, but I'm not sure I entirely get it...

So, what happens if I integrate (3x-5)^4 using int((ax + b)dx)=1/2ax^2+bx...

I get 3/2x^2+5x??? That's not right, I can't be understanding your notation or something.

 

[Aelus]

<< So, what happens if I integrate (3x-5)^4 using int((ax + b)dx)=1/2ax^2+bx... >>



int((3x-5)^4)dx isn't exactly the easiest of integrals, unless i'm not seeing the better way, your best bet is to solve the (3x-5)^4 so you have separate pieces, it's easy to integrate 3x-5, because you just integrate 3x, and then you integrate -5, but it's a pain to integrate functions which are multiplications of itself. hope that makes things a bit clearer.

Aelus

 

[JayHu]

okay i didnt even read your post correctly.. oops.
i just wanted to correct the person above me.

so getting to your quesiton of integral (3x-5)^4

i think your best bet is to expand it and go from there..
there really isnt any easy way to do that. unless im missing something.
it doesnt take too long to expand it via binomial theorem.

 

[Dougster]

Thanks for your help, but I'm fairly sure that's not the easiest way.

I just can't put my finger on the "reverse" formula... but through trial and error I was able to work out that int((3x-5)^4)dx = 1/15(3x-5)^5 + c

It make sense if you thing about it...

D((ax + b)^n)= na(ax+b)^(n-1)

so

D(1/15(3x-5)^5)= (3x-5)^4)

(1/15 * 3 * 5 = 1 so coefficient of bracket is removed)

 

[Geekbabe]

OMG,That's it.. I gotta start wearing my glasses I read the subject of this thread really quick and thought it said


"is there a man boffin out there who can help me quickie "

 

[Dougster]

<< okay i didnt even read your post correctly.. oops.
i just wanted to correct the person above me.
>>

Sorry! I just re-read my post and now see how it looks, I wasn't having a go at you, my "I know that..." wasn't directed at you, it was directed at br0wn and was backing you up. How arrogant does that look! Apologies all round.

 

[Dougster]

Whoops, just noticed an error in my first post, it should have read "integral of (ax + b)^n?" not "integral of (ax + b). I have corrected.

 

[Dougster]

Ok I think I have it... but am I right? Can someone poke a hole in it etc.

int((ax+b)^n) = (1/(a(n+1)))(ax+b)^(n+1)

So with int((3x-5)^4) = (1/(3*5))(3x-5)^5) = 1/15(3x-5)^5

That looks so much neater written down on paper!

 

[JayHu]

ACK!
you are soo right.
thats good.
i've been away from high school calc for too long.
im starting to think interms of harder equations..
but yeah i see that.
i looks to be right. but to be honest i was looking at (ax^m+b)^n instead of the case of m=1.. oops.

but looks good. wtg!