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Integration - The Basics



Hi,

I understand how to get to the answer, that's fine. But I just don't understand the logic used. "Consider", what do they mean? Are we just "Letting" y=... or is y=sin(2x+3) the actual integral?

If so, if we differentiate this value, we should get back to cos(2x+3) but we clearly don't. So this value cannot be the integral? Why are we taking these steps?

Thanks!

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Original post by ps1265A


Hi,

I understand how to get to the answer, that's fine. But I just don't understand the logic used. "Consider", what do they mean? Are we just "Letting" y=... or is y=sin(2x+3) the actual integral?

If so, if we differentiate this value, we should get back to cos(2x+3) but we clearly don't. So this value cannot be the integral? Why are we taking these steps?

Thanks!


The answer is y=1/2 sin(2x+3)

If you differentiate that you get cos(2x+3)

The text is explaining why you you need the 1/2
Reply 2
Original post by TenOfThem
The answer is y=1/2 sin(2x+3)

If you differentiate that you get cos(2x+3)

The text is explaining why you you need the 1/2


I want to know whether step 1 is just a "rough" integration


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Original post by ps1265A
I want to know whether step 1 is just a "rough" integration


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No

Step 1 is telling you the thinking process
Reply 4
Original post by TenOfThem
No

Step 1 is telling you the thinking process


And what is the thinking process? An estimation?


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Original post by ps1265A
And what is the thinking process? An estimation?


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It is looking at the integrated and saying to yourself ... Oh ... I must have differentiated a Sin function in order to get a Cos function


I am struggling to see what is confusing you
Original post by ps1265A
I want to know whether step 1 is just a "rough" integration


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Step 1 is an "intelligent " guess. You then check whether it is correct and adjust if necessary.
You should try to get to know certain standard patterns. For example, consider a function of x f(x). If the integral is F(x) then the integral of f(ax) will be F(ax)/a
Similarly the integral of f(ax+b) will be F(ax+b)/a.
IN reverse, if the derivative of f(x) is g(x) say then the derivative of f(ax) will be ag(ax) and of f(ax+b) will be ag(ax+b).
More generally the derivative of f(g(x)) will be g'(x)f'(g(x))
See page 38 in the attachment.
Reply 7
Original post by TenOfThem
It is looking at the integrated and saying to yourself ... Oh ... I must have differentiated a Sin function in order to get a Cos function


I am struggling to see what is confusing you


Yes, that's exactly what I wanted to know and was thinking the same, just didn't know how to word it :smile:


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Reply 8
Original post by brianeverit
Step 1 is an "intelligent " guess. You then check whether it is correct and adjust if necessary.
You should try to get to know certain standard patterns. For example, consider a function of x f(x). If the integral is F(x) then the integral of f(ax) will be F(ax)/a
Similarly the integral of f(ax+b) will be F(ax+b)/a.
IN reverse, if the derivative of f(x) is g(x) say then the derivative of f(ax) will be ag(ax) and of f(ax+b) will be ag(ax+b).
More generally the derivative of f(g(x)) will be g'(x)f'(g(x))
See page 38 in the attachment.


Thank you soooo much, I understand it now :smile:


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Original post by ps1265A
Thank you soooo much, I understand it now :smile:


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You''re welcome
Reply 10
Original post by brianeverit
You''re welcome


Could you tell me what the best way to learn integration is? At the moment, I have 2 option: one is by simply looking at the formulas and just plugging in and one is by actually doing the steps as I did above. Or do I have to adopt a combination of both because all questions don't follow the above method?


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(edited 9 years ago)
Reply 11
Original post by TenOfThem
It is looking at the integrated and saying to yourself ... Oh ... I must have differentiated a Sin function in order to get a Cos function


I am struggling to see what is confusing you


Could you tell me what the best way to learn integration is? At the moment, I have 2 option: one is by simply looking at the formulas and just plugging in and one is by actually doing the steps as I did above. Or do I have to adopt a combination of both because all questions don't follow the above method?


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Reply 12
Original post by brianeverit
You''re welcome


Because say if I have the Q was integrate 1/(2x+1)^2 and I followed the 3 step method, I would get:
ImageUploadedByStudent Room1421517816.692358.jpg

But the actual answer is -1/2(2x+1)^2


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Original post by ps1265A
Because say if I have the Q was integrate 1/(2x+1)^2 and I followed the 3 step method, I would get:
ImageUploadedByStudent Room1421517816.692358.jpg

But the actual answer is -1/2(2x+1)^2


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Both this question and the original question use the idea of "reverse chain rule"

You should know that differentiating the function of a function needs the chain rule

So if you are integrating the function of a function you will consider reversing the chain rule
Reply 14
Original post by TenOfThem
Both this question and the original question use the idea of "reverse chain rule"

You should know that differentiating the function of a function needs the chain rule

So if you are integrating the function of a function you will consider reversing the chain rule


I have used the reverse chain rules


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Original post by ps1265A
I have used the reverse chain rules


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Sorry did not look at your work properly

You need to increase the power

You have changed a power of -2 to a power of -3

That is decreasing
(edited 9 years ago)
Reply 16
Original post by TenOfThem
Sorry did not look at your work properly

You need to increase the power

You have changed a power of -2 to a power of -3

That is decreasing


I'm sorry if the picture is unclear. I've changed my power from a 2 to a 3


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Reply 17
Original post by TenOfThem
Sorry did not look at your work properly

You need to increase the power

You have changed a power of -2 to a power of -3

That is decreasing


Or do I have to account for the reciprocal?


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Original post by ps1265A
Or do I have to account for the reciprocal?


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I can see that you have changed from a 2to a 3

As you have realised now ... You have changed -2 to -3 ... Decreasing rather than increasing ... Since the function is in the denominator
Reply 19
Original post by TenOfThem
I can see that you have changed from a 2to a 3

As you have realised now ... You have changed -2 to -3 ... Decreasing rather than increasing ... Since the function is in the denominator


So this is what I've got:

ImageUploadedByStudent Room1421520876.060318.jpg
ImageUploadedByStudent Room1421520911.005440.jpg


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