Integration of 3^(x-1) (1 Viewer)

[lyounamu]

How do you integrate 3^(x-1)?

 

[tommykins]

3^[x-1]/ln3

I forgot how, but I know that's the answer (or should be? )

 

[Captin gay]

ues the fact that 3^(x-1) = e^(ln (3^(x-1)))

log both sides to see why

 

[vds700]

How do you integrate 3^(x-1)?

let y = 3^x
x = log₃y
=lny/ln3
dx/dy = 1/yln3
dy/dx = yln 3
=3^x ln3
there4fore I(3^x ln3) dx = 3^x

I3^x-1 dx
=I3^x . (1/3) dx
=(1/3) 3^x /ln3 (from above)
=3^x-1/ln3

hope this makes sense

 

[lyounamu]

let y = 3^x
x = log₃y
=lny/ln3
dx/dy = 1/yln3
dy/dx = yln 3
=3^x ln3
there4fore I(3^x ln3) dx = 3^x

I3^x-1 dx
=I3^x . (1/3) dx
=(1/3) 3^x /ln3 (from above)
=3^x-1/ln3

hope this makes sense

Very comprehensive, Andrew. Appreciate it.

Thanks to other posts above.

 

[morning storm]

haha you guys are good.

ioh and namu, did you know i made the video in your signature haha...

 

[lyounamu]

haha you guys are good.

ioh and namu, did you know i made the video in your signature haha...

of course, i knew that. That's an awesome video, man. Pity that you weren't there to show your stuff. Notable participants were only like Andrew (not vds) and fusca.

 

[Slidey]

How do you integrate 3^(x-1)?

Sometimes when you don't know how to integrate, it helps to derive them if they're simple functions. ESPECIALLY true of exponentials.

Derivative: (x-1)' * ln(3) * 3^(x-1) = ln(3) * 3^(x-1).

Thus the integral would probably be the opposite, or 3^(x-1)/ln(3).

I say this because in a test when you forget, it really helps to know ways to 'guess' the answer with a high degree of accuracy and speed.

-----------------------------------------------------

If I were vds, I'd be showing you how to derive it using this method, personally (implicit differentiation - 4 unit technique, but a great tool to learn, and very easy, and clearer):

y=3^(x-1)
Convert x to a function of y using the natural logarithmic function:
ln(y) = (x-1)*ln(3)
Take derivative w.r.t. x:
d(ln(y))/dx = ln(3)*(x-1)/dx
y'/y = ln(3) (remember d(ln(f(x)))/dx = f'(x)/f(x) because y is a function of x - chain rule)
y'=ln(3)y
y'=ln(3)*3^(x-1)

or shorter:
y=3^(x-1)
ln(y) = ln(3)*(x-1)
y'/y = ln(3)
y'=ln(3)*y
y'=ln(3)*3^(x-1)

Although, vds is actually implicitly using implicit differentiation anyway, which is a bit of a mind fuck - because they're actually both equivalent to the chain rule. The benefit to his method I guess is that I think you get taught his 'derivative of an inverse function' method in normal 2 unit maths.

Also, if you know the integral of 3^x but not 3^(x-1), use 'integration by substitution' (a 3 unit technique which is the reverse chain rule)).

let u=x-1
then d(u)/dx = d(x-1)/dx = 1
but it's easier to just go: du = 1 dr, or du=dx
thus y=3^u
hence integral Int 3^(x-1) dx becomes Int 3^u du
answer: 3^u/ln(3)
sub in u=x-1:
final answer: Int y = 3^(x-1)/ln(3)

it's based on the fact that: dy/dx = du/dx * dy/du (chain rule again).

E.g.:
chain rule:
y=(x^2+1)^7
let u=x^2+1, u'=2x
y=u^7
y'=7*u'*u^6 (u is a function of x)
y' = 14x(x^2+1)^6
Or more formally:
dy/dx = dy/du * du/dx
dy/dx = 7u^6 * u' = 7(x^2+1)^6 * 2x = 14x(x^2+1)^6

inverse chain rule:
y = 14x(x^2+1)^6
Int y dx = Int 14x(x^2+1)^6 dx
let u = x^2+1, du/dx = 2x, du = 2x dx
so Int 14x(x^2+1)^6 dx = Int 7 * u^6 * du
answer = u^7 = (x^2+1)^7

short version:
Int 14x(x^2+1)^6 dx
u=x^2+1, du=2x dx
Int 7u^6 dx = u^7
= (x^2+1)^7

 

[lolokay]

yeah.. or you could just

ues the fact that 3^(x-1) = e^(ln (3^(x-1)))

 

[Slidey]

If you'd read my post, it's about understanding multiple methods. This is important, because Iyounamu is doing Maths Extension 2, where maths means a little more than simply rote learning.

But hey, you'll go real far with that attitude: "This method works, thus I don't need to understand why or bother with alternate methods."

I'm not sure why I didn't put it in the forum guidelines, but just so you know, multiple solutions to a problem are encouraged here for precisely the reasons above. It's not just about solving somebody's homework problem and moving on.

 

[vds700]

haha you guys are good.

ioh and namu, did you know i made the video in your signature haha...

HEY TOM!! finally found u on BOS. lol.

 

[vds700]

Sometimes when you don't know how to integrate, it helps to derive them if they're simple functions. ESPECIALLY true of exponentials.

Derivative: (x-1)' * ln(3) * 3^(x-1) = ln(3) * 3^(x-1).

Thus the integral would probably be the opposite, or 3^(x-1)/ln(3).

I say this because in a test when you forget, it really helps to know ways to 'guess' the answer with a high degree of accuracy and speed.

-----------------------------------------------------

If I were vds, I'd be showing you how to derive it using this method, personally (implicit differentiation - 4 unit technique, but a great tool to learn, and very easy, and clearer):

y=3^(x-1)
Convert x to a function of y using the natural logarithmic function:
ln(y) = (x-1)*ln(3)
Take derivative w.r.t. x:
d(ln(y))/dx = ln(3)*(x-1)/dx
y'/y = ln(3) (remember d(ln(f(x)))/dx = f'(x)/f(x) because y is a function of x - chain rule)
y'=ln(3)y
y'=ln(3)*3^(x-1)

or shorter:
y=3^(x-1)
ln(y) = ln(3)*(x-1)
y'/y = ln(3)
y'=ln(3)*y
y'=ln(3)*3^(x-1)

Although, vds is actually implicitly using implicit differentiation anyway, which is a bit of a mind fuck - because they're actually both equivalent to the chain rule. The benefit to his method I guess is that I think you get taught his 'derivative of an inverse function' method in normal 2 unit maths.

Also, if you know the integral of 3^x but not 3^(x-1), use 'integration by substitution' (a 3 unit technique which is the reverse chain rule)).

let u=x-1
then d(u)/dx = d(x-1)/dx = 1
but it's easier to just go: du = 1 dr, or du=dx
thus y=3^u
hence integral Int 3^(x-1) dx becomes Int 3^u du
answer: 3^u/ln(3)
sub in u=x-1:
final answer: Int y = 3^(x-1)/ln(3)

it's based on the fact that: dy/dx = du/dx * dy/du (chain rule again).

E.g.:
chain rule:
y=(x^2+1)^7
let u=x^2+1, u'=2x
y=u^7
y'=7*u'*u^6 (u is a function of x)
y' = 14x(x^2+1)^6
Or more formally:
dy/dx = dy/du * du/dx
dy/dx = 7u^6 * u' = 7(x^2+1)^6 * 2x = 14x(x^2+1)^6

inverse chain rule:
y = 14x(x^2+1)^6
Int y dx = Int 14x(x^2+1)^6 dx
let u = x^2+1, du/dx = 2x, du = 2x dx
so Int 14x(x^2+1)^6 dx = Int 7 * u^6 * du
answer = u^7 = (x^2+1)^7

short version:
Int 14x(x^2+1)^6 dx
u=x^2+1, du=2x dx
Int 7u^6 dx = u^7
= (x^2+1)^7

nice work mate- I didn't think to use implicit differentiation.

 

[lyounamu]

If you'd read my post, it's about understanding multiple methods. This is important, because Iyounamu is doing Maths Extension 2, where maths means a little more than simply rote learning.

But hey, you'll go real far with that attitude: "This method works, thus I don't need to understand why or bother with alternate methods."

I'm not sure why I didn't put it in the forum guidelines, but just so you know, multiple solutions to a problem are encouraged here for precisely the reasons above. It's not just about solving somebody's homework problem and moving on.

You are awesome! Seriously, you are not just there to help someone. You are there to teach someone a valuable lesson too! I learnt heaps from you from this thread. Thanks.

Thanks to other posters here.

To vds: hahahaha...i was aware of his membership for a while now.