Integrate 3^x (1 Viewer)

[overwhelming]

how do we do this?

 

[~*HSC 4 life*~]

ok quick derivation:
generally is
y= a^x
then x=log(base a)y
x= lny/lna (change of base rule

dx/dy= 1/lna x 1/y
dy/dx= ylna
= a^x * lna

so to integrate 3^x you get 3^x * ln3

 

[Heinz]

actually i think the answer is 3/ln3 + c

 

[~*HSC 4 life*~]

oh...
lol
i dont get whats wrogn with my way

 

[Heinz]

wait, i did it again and got (3^x)/ln3 +c

 

[CM_Tutor]

Heinz is right, the answer is 3^x / ln 3 + C, and in general

derivative of A^x is A^x . ln A (for the reason shown by ~*HSC 4 life*~)
integral of A^x is A^x / ln A + C.

~*HSC 4 life*~, your error is in the last step. You have shown that if y = 3^x, then dy / dx = 3^x . ln 3. Integrate this with respect to x and you will get
y = int (3^x . ln 3) dx
Now notice that y = 3^x, and multiplying both sides by 1 / ln 3, we get:
int 3^x dx = 3^x / ln 3 + C, for some constant C.

 

[crazylilmonkee]

i cant believe i forgotten all this stuff already

 

[~*HSC 4 life*~]

ooh SUGAR!

sorry...i misread the q and found the derivative and not the integral lol

 

[Xayma]

Originally posted by CM_Tutor
Heinz is right, the answer is 3^x / ln 3 + C, and in general

derivative of A^x is A^x .

Itsn't the derivative of A^x=kA^x where k is a constant (where it equals 1 for e)

 

[Heinz]

Originally posted by Xayma
Itsn't the derivative of A^x=kA^x where k is a constant (where it equals 1 for e)

The derivative of A^kx = kA^kx but here k = 1 so d/dx of A^x = A^x. What do you mean by "where it equals 1 for e"?

 

[CM_Tutor]

No, the derivative of A^(kx) is A^(kx) . k . ln A

Proof: Let y = A^(kx)
Then y = e^(ln A^(kx)) = e^(kx . ln A)
Now, the derivate with respect to x of e^(Cx) is Ce^(Cx) where C is some constant.
And here C = kln A, and so dy/dx = kln A . e^(kx . ln A) = kln A . y = A^(kx) . k . ln A

NOTE TO Xayma, you have misquoted me, as the dot is a multiplication sign (maybe I should use * from now on.) What I said was that, in general, the derivative of A^x is A^x . ln A (ie. A^x times ln A). Your misquote, that the derivative of A^x is A^x is only true if A = e.

Your statement that "Itsn't the derivative of A^x=kA^x where k is a constant (where it equals 1 for e)" is correct provided the constant k is ln A.

 

[Heinz]

Originally posted by CM_Tutor
No, the derivative of A^(kx) is A^(kx) . k . ln A

Oops, i thought A was e. Because i was thinking of Ae^kt ... and theyre so close together...all looks the same to me. my bad ^^;

 

[overwhelming]

so what is the final answer lol

 

[CM_Tutor]

The final answer is that the integral of 3^x is 3^x / ln 3 + C, for some constant C.

 

[Grey Council]

Tell James (lazarus) to make your status "teacher". That way people won't argue with you etc.

not that i'm saying anyone has argued with you. I'm just saying.

it looks cool as well.

 

[Xayma]

Originally posted by Heinz
What do you mean by "where it equals 1 for e"?

The constant (k) =1 when the number is e ie the derivative of e^x=e^x
If the number is smaller then e (eg 2^x k will be smaller) If it is bigger than e k will be bigger.

Originally posted by CM_Tutor
NOTE TO Xayma, you have misquoted me, as the dot is a multiplication sign (maybe I should use * from now on.) What I said was that, in general, the derivative of A^x is A^x . ln A (ie. A^x times ln A). Your misquote, that the derivative of A^x is A^x is only true if A = e.

Its the space that through me off I thought you were starting a new sentance. Sorry

 

[overwhelming]

Originally posted by CM_Tutor
The final answer is that the integral of 3^x is 3^x / ln 3 + C, for some constant C.

do u mean (3^x/ ln 3) + C?

 

[CM_Tutor]

Yes, I mean (3^x / ln 3) + C, or written another way C + 3^x / ln 3.

Note that what I have written, 3^x / ln3 + C is correct, as that does not imply that C is on the denominator. For that, I would need to have written 3^x / (ln 3 + C) - Hope that clears up your confusion.

 

[KeypadSDM]

Originally posted by CM_Tutor
Note that what I have written, 3^x / ln3 + C is correct, as that does not imply that C is on the denominator. For that, I would need to have written 3^x / (ln 3 + C) - Hope that clears up your confusion.

Always put everything in brackets on this forum. Most people don't understand order of operations and tend to put anything added or subtracted after the fraction on the denominator.

 

[CM_Tutor]

I'll bear that in mind. Thanks.