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

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