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anonymous
 one year ago
Integrals
anonymous
 one year ago
Integrals

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rational
 one year ago
Best ResponseYou've already chosen the best response.3show that the derivative is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I did was split the variables up so instead of \[\Large F(x)=\int\limits_{x}^{3x}\frac{dt}{t}\] I hose 2 as a constant and turned into \[\Large F(x)=\int\limits\limits_{x}^{2}\frac{dt}{t}+\int\limits\limits_{2}^{3x}\frac{dt}{t}\] \[\Large F(x)=\int\limits\limits\limits_{2}^{x}\frac{dt}{t}+\int\limits\limits\limits_{2}^{3x}\frac{dt}{t}\] And now F'(x)=f(x) \[\Large F'(x)=\frac{ 1 }{ x }+\frac{ 1 }{ 3x }=\frac{ 2 }{ 3x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thatv doesn't look like 0, unless I was going about it the wrong way

rational
 one year ago
Best ResponseYou've already chosen the best response.3you forgot to use chain rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I thought it was 1/3 x, I see

rational
 one year ago
Best ResponseYou've already chosen the best response.3\[\dfrac{d}{dx}\int\limits_a^{g(x)} ~f(t)~dt = f(g(x))*\color{red}{g'(x)}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1is it not that int of 1/t dt = lnt? plug limits in, we have ln 3x  ln x = ln 3 constant

rational
 one year ago
Best ResponseYou've already chosen the best response.3that looks nice and more direct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Loser66 That makes it so much easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll give my medal to @rational since he helped me first

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1hihihi... my brain can't think of something complicated.
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