anonymous one year ago Integrals

1. anonymous

2. rational

show that the derivative is 0

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

4. anonymous

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

5. rational

you forgot to use chain rule

6. anonymous

Where?

7. anonymous

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

8. rational

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

9. anonymous

Oh, it's now 0

10. Loser66

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

11. rational

that looks nice and more direct

12. anonymous

@Loser66 That makes it so much easier

13. anonymous

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

14. Loser66

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