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anonymous
 one year ago
How would I do this?
anonymous
 one year ago
How would I do this?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... but I don't have the function... Im kind of confused

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\int\limits_0^2 f(2x)\,dx~~\stackrel{u=2x}{=}~~ \frac{1}{2}\int\limits_0^4 f(u)\,du = \frac{1}{2}*20=10\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why does the limit of integration change from 4 to 2?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4scratch that, lets work it again from beginning

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you want to evaluate \[\int\limits_0^2f(2x)\, dx\] substitute \(u=2x \implies du=2dx\implies dx=\dfrac{du}{2}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4next work the bounds, as \(x\to 0\), what does \(u\to\) ? as \(x\to 2\), what does \(u\to\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01) x > 0, u > 0 2) x > 2, u > 4

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4so upon substitution, bounds change from (0, 2) to (0, 4) and the differential changes from dx to du/2 plug them in

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, so then that's \[\int\limits_{0}^{4}f(u) \frac{ du }{ 2 }\] and then the 1/2 comes out of the integral

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Yes, next recall that the variable in definite integral is "dummy" \[\int\limits_a^b f(\color{red}{x})\,d\color{red}{x} = \int\limits_a^b f(\color{red}{t})\,d\color{red}{t}=\int\limits_a^b f(\color{red}{\spadesuit})\,d\color{red}{\spadesuit}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh now I get what's happening!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Thanks so much!
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