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integration of lnx (1 Viewer)

_ShiFTy_

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You will not need to learn how to integrate lnx in 2unit..

However, you will need to use 'by parts' to integrate lnx (4unit)
 

darkliight

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Remember, when you integrate a function you can think of it this way "what function gives me this one when I differentiate it?". Basically, it's the opposite of differentiation.

For example, integral x^2 dx. What function gives me x^2 when I differentiate it? It doesn't take long to realise that x^3/3 works (differentiate it and check if you want).

We can think the same way when trying to integrate ln(x). What function differentiates to give me ln(x)? Well, we don't really know a single function that differentiates to ln(x) ... but what about x*ln(x)? The derivative of this, using the product rule, gives us

x*1/x + 1*ln(x)
= 1 + ln(x).

Now, that's pretty close .... all except that extra 1. Can you think of what else we would have to add to x*ln(x) so that when we differentiate, we get the 1 to cancel?
 
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insert-username

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Following on from darkliight, the integral is:

x.loge(x) - x

But don't worry. In 2 unit they will guide you to it, i.e. something along the lines of:

Differentiate [x.ln(x) - x]. Hence or otherwise, evaluate 1∫2logex dx


I_F
 
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pLuvia

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In 2 unit and 3 unit maths you could be asked to integrate lnx but using trapezoidal rule or simpsons rule, but a direct integration, that would need integration by parts
 

Riviet

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insert-username said:
Differentiate [x.ln(x) - x]. Hence or otherwise, evaluate 1∫2logex dx
These kind of questions are quite common in 2/3 unit exams, so look out for these and get lots of practice in preparation.
 
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onebytwo

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if this helps
if your asked to integrate a definite integral, then try and relate the lnx curve to the exponential curve. use the area of a rectangle minus the area to the left of the curve......if that makes sense
 

onebytwo

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if this helps
if your asked to integrate a definite integral, then try and relate the lnx curve to the exponential curve. use the area of a rectangle minus the area to the left of the curve......if that makes sense
 

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