differentiation (1 Viewer)

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let u = 1 - x^2

du = -2x dx

don't forget to change limits! you can check this if you like but its from 1--> 3/4

to cancel out the -2 in front of the dx multiply the integral by -1/2

-1/2 * integral (du/u), limits 1-->3/4

= 1/2 * integral (du/u), limits 3/4-->1

= 1/2 ln(u), limits 3/4-->1

= 1/2 * [ ln(1)- ln (3/4) ]

= 1/2 * [ ln(1/ {3/4} ) ]

= 1/2 * [ ln(4/3) ]
 

bos1234

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why do you have to do this?

to cancel out the -2 in front of the dx multiply the integral by -1/2
 
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bos1234 said:
why do you have to do this?

to cancel out the -2 in front of the dx multiply the integral by -1/2
hmm i thought that bit was always going to be hard to convey on a forum. sorry about that.

du = -2x dx

x dx = -du/2 is a better way of putting it.

if you sub -du/2 for x dx and u for 1 - x^2 you get

integral (-du/2u), limits 1-->3/4

or simply -1/2 * integral (du/u), limits 1-->3/4
 

elseany

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when the numerator is a degree lower than the denominator you should instantly consider logs. the trick with these questions is that you need to get the top looking like the derivative of the bottom, if you can recognise that and manipulate the top you can skip the process of substitution.

so in this question to get the top looking like the derivative of the bottom you gotta multiply by -2, so to balance this you need to multiply by a -1/2 outside the integral.
 

bos1234

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elseany said:
when the numerator is a degree lower than the denominator you should instantly consider logs. the trick with these questions is that you need to get the top looking like the derivative of the bottom, if you can recognise that and manipulate the top you can skip the process of substitution.

so in this question to get the top looking like the derivative of the bottom you gotta multiply by -2, so to balance this you need to multiply by a -1/2 outside the integral.
thanks for thius tip
 

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