Integral question

  • Thread starter evan4888
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Here is the problem:

[tex] \int ^2_1 \frac{dx}{(3-5x)^2} [/tex]

[tex] u = 3-5x [/tex]

[tex] du = -5dx [/tex]

[tex] dx = -\frac{1}{5} du [/tex]

so, (with the 7 and 2 being negatives)

= [tex] -\frac{1}{5} \int^7_2 \frac{du}{u^2} [/tex]

= [tex] -\frac{1}{5} (-\frac{1}{u})]^7_2 [/tex]

But what I don't understand is how the [tex] u^2 [/tex] becomes just [tex] u [/tex].
 
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Why did you change your bounds? You had everything right.

[tex]\frac{1}{u^2}[/tex] can be written as [tex]u^{-2}[/tex]

Now follow the general power rule and [tex]\int u^{-2}du = \frac{(u^{-1})}{(-1)}[/tex]

But to write that more nicely write it as [tex]-\frac{1}{u}[/tex]

Make sense?
 
Last edited:

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