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help please (integration)

Ok so we look at integration by taking logarithms then integration but I keep getting the wrong answer?


[br]dy/dx=2x[br]ln(dy/dx)=ln(2x)[br]ln(dy)−ln(dx)=xln(2)[br]∫ln(dy)−ln(dx)dx=(x2/2)ln(2)[br]1/2(ln(dy))2+1/2(ln(dx))2=(x2/2)ln(2)[br][br]dy/dx=2^x[br]ln(dy/dx)=ln(2^x)[br]ln(dy)-ln(dx)=xln(2)[br]\int ln(dy)-ln(dx) dx=(x^2/2)ln(2)[br]1/2 (ln(dy))^2 + 1/2 (ln(dx))^2=(x^2/2)ln(2)[br]

How do I get rid of the dy's and dx's they do not cancel:confused:

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If you are Integrating Dy/Dx, use by parts, Make U=x and Dv/Dx=2^u
So Du/Dx=1 and v=2^u, then sub into the formula
Original post by user6
Ok so we look at integration by taking logarithms then integration but I keep getting the wrong answer?


[br]dy/dx=2x[br]ln(dy/dx)=ln(2x)[br]ln(dy)−ln(dx)=xln(2)[br]∫ln(dy)−ln(dx)dx=(x2/2)ln(2)[br]1/2(ln(dy))2+1/2(ln(dx))2=(x2/2)ln(2)[br][br]dy/dx=2^x[br]ln(dy/dx)=ln(2^x)[br]ln(dy)-ln(dx)=xln(2)[br]\int ln(dy)-ln(dx) dx=(x^2/2)ln(2)[br]1/2 (ln(dy))^2 + 1/2 (ln(dx))^2=(x^2/2)ln(2)[br]

How do I get rid of the dy's and dx's they do not cancel:confused:


dydx\frac{dy}{dx} isn't really a fraction is such, it's just the notation used to represent the derivative of y with respect to x.

It may help you if you write 2x=exln2.2^x = e^{xln2}.
Reply 3
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user6
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Original post by brittanna
dydx\frac{dy}{dx} isn't really a fraction is such, it's just the notation used to represent the derivative of y with respect to x.

It may help you if you write 2x=exln2.2^x = e^{xln2}.


so
2x=exln2[br]dy/dx=(ln2)exln2[br]2^x=e^{xln2}[br]dy/dx=(ln2)e^{xln2}[br] ?
Original post by brittanna
dydx\frac{dy}{dx} isn't really a fraction is such, it's just the notation used to represent the derivative of y with respect to x.

It may help you if you write 2x=exln2.2^x = e^{xln2}.


dy/dx is a fraction this method wouldn't work though because for a start dy and dx tend towards zero so division by 0 so you can't separate the logarithms that way, hence why you have to take limits when evaluating dy/dx because its 0/0.
Original post by user6
so
2x=exln2[br]dy/dx=(ln2)exln2[br]2^x=e^{xln2}[br]dy/dx=(ln2)e^{xln2}[br] ?


If we had y=exln2y=e^{xln2} then that would be true.

But we have dydx=2x=exln2\frac{dy}{dx} = 2^x = e^{xln2}.

So integrating, we get y=∫exln2dx y = \int e^{xln2} dx.

Can you evaluate this integral?
Reply 6
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user6
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Original post by brittanna
If we had y=exln2y=e^{xln2} then that would be true.

But we have dydx=2x=exln2\frac{dy}{dx} = 2^x = e^{xln2}.

So integrating, we get y=∫exln2dx y = \int e^{xln2} dx.

Can you evaluate this integral?


∫exln2dx=(1/2)(exln2)2ln2\int e^{xln2} dx= \frac{(1/2)(e^{xln2})^2}{ln2}
Original post by user6
∫exln2dx=(1/2)(exln2)2ln2\int e^{xln2} dx= \frac{(1/2)(e^{xln2})^2}{ln2}


Not quite. What do you get if you integrate exe^x?

What about eaxe^{ax} where a is a real number?

(You can try differentiating it afterwards to see if you get back to what you started with).
Reply 8
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Original post by brittanna
Not quite. What do you get if you integrate exe^x?

What about eaxe^{ax} where a is a real number?

(You can try differentiating it afterwards to see if you get back to what you started with).


integrate of e^x is = e(1/2)(x2)e^{(1/2)(x^2)}
Original post by Dalek1099
dy/dx is a fraction this method wouldn't work though because for a start dy and dx tend towards zero so division by 0 so you can't separate the logarithms that way, hence why you have to take limits when evaluating dy/dx because its 0/0.


dydx\frac{dy}{dx} isn't a fraction. It represents a single limit.
Original post by brittanna
dydx\frac{dy}{dx} isn't a fraction. It represents a single limit.


If you do M5 you will realise this not true when you have to divide by dx or dt depending on the question you could look it up.
(edited 9 years ago)
Reply 11
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Rkai01
6
Woah, what unit is this for?
Is it FP2 or FP3?
Original post by user6
integrate of e^x is = e(1/2)(x2)e^{(1/2)(x^2)}


The integral of exe^x is e^x. Have you covered this in class yet?
Original post by Rkai01
Woah, what unit is this for?
Is it FP2 or FP3?


It often comes up in C3/ C4 ...
Reply 14
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user6
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Original post by brittanna
The integral of exe^x is e^x. Have you covered this in class yet?


yes sorry i forgot
Original post by Dalek1099
If you do M5 you will realise this not true when you have to divide by dx or dt depending on the question you could look it up.


I'm at university, but thanks for letting me know...
Original post by brittanna
I'm at university, but thanks for letting me know...


I have only really seen dy/dx treated as a fraction in that topic but it can be implied from other knowledge that dy/dx is basically the division of two infinitesimally small quantities and its how they compare to each other that determines its value, in M5 you divide by dt and dx and take all the limits as all the dt,dx expressions approach zero and you often get expressions like 4dx or 5dxdx/dt and the understanding that both of these approach zero can only be understood by taking both limits separately.It is only when looking at this M5 topic did I truly appreciate that it was a fraction but I thought it was based on other rules that obey the law of fractions.
(edited 9 years ago)
Original post by user6
yes sorry i forgot


So do you know how to integrate eaxnowe^{ax} now (and hence exln2e^{xln2})?
Reply 18
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Rkai01
6
Original post by brittanna
It often comes up in C3/ C4 ...


What? I have already done those mods! Never came up. C3 there is no integrating whatsoever.
This is for edexcel right?
Reply 19
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Rkai01
6
Original post by brittanna
So do you know how to integrate eaxnowe^{ax} now (and hence exln2e^{xln2})?


From what I remember I think it just becomes integrating 2^x

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