Integrals

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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show that the derivative is 0
What 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) \[\Large F'(x)=-\frac{ 1 }{ x }+\frac{ 1 }{ 3x }=\frac{ 2 }{ 3x }\]

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Thatv doesn't look like 0, unless I was going about it the wrong way
you forgot to use chain rule
Where?
Oh, I thought it was 1/3 x, I see
\[\dfrac{d}{dx}\int\limits_a^{g(x)} ~f(t)~dt = f(g(x))*\color{red}{g'(x)}\]
Oh, it's now 0
is it not that int of 1/t dt = lnt? plug limits in, we have ln 3x - ln x = ln 3 constant
that looks nice and more direct
@Loser66 That makes it so much easier
I'll give my medal to @rational since he helped me first
hihihi... my brain can't think of something complicated.

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