Question

Integrals

Answer 1

show that the derivative is 0

Answer 2

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)
\[\Large F'(x)=-\frac{ 1 }{ x }+\frac{ 1 }{ 3x }=\frac{ 2 }{ 3x }\]

Answer 3

Thatv doesn't look like 0, unless I was going about it the wrong way

Answer 4

you forgot to use chain rule

Answer 5

Where?

Answer 6

Oh, I thought it was 1/3 x, I see

Answer 7

\[\dfrac{d}{dx}\int\limits_a^{g(x)} ~f(t)~dt = f(g(x))*\color{red}{g'(x)}\]

Answer 8

Oh, it's now 0

Answer 9

is it not that int of 1/t dt = lnt? plug limits in, we have ln 3x - ln x = ln 3 constant

Answer 10

that looks nice and more direct

Answer 11

@Loser66 That makes it so much easier

Answer 12

I'll give my medal to @rational since he helped me first

Answer 13

hihihi... my brain can't think of something complicated.