## anonymous 3 years ago A particle starts at x(0) = 2. If its velocity is given by v(t) = ln(1+t), find its position at t = 5

1. anonymous

Take the integral of velocity from 0 to 5 and then add 2. $\int\limits_{0}^{5} \ln(1+t) dt$ Set u = 1+t, du = dt So you have $\int\limits_{0}^{5}\ln(u)du$ Which becomes [uln(u) - u] from 0 to 5 Plug in your values to get 5(ln(5) - 1) Then add 2, so the position at t =5 is 2 + 5(ln(5) -1)

2. anonymous

is there a way to find it without u substitution?

3. anonymous

Nope. It's the only way to do that sort of integral unfortunately.

4. anonymous

really? oh I thought you can use it with FTC

5. anonymous

The FTC states that if you have a function F(x) which is the integral of another function G(x) then the derivative of F(x) is simply G(x). Additionally, the integral of G(x) is just F(b) - F(a).

6. anonymous

oh kinda like if it's F'(x) = f(x)?

7. anonymous

Pretty much.

8. anonymous

ok thanks :)

9. anonymous

but wait, isn't u supposed to be ln (u) and du is 1+t? because when i plug in your answer, there is no answer choices for it :/

10. anonymous

11. anonymous

Ugh, sorry, it's been a long day. (ulnu - u) from 0 to 5 should be (1 + t)(ln(1+t) - 1) so it's actually 6(ln6 -1).

12. anonymous

Because u was substituted for 1+t earlier, so we need to exchange it back out.

13. anonymous

i keep getting 9.65 but it is not an answer choice? i think there's a step missing or a miscalculation :( the answer choices are : a) 5.751 b) 7.751 c) 1.792 d) 3.792 e) 3.751

14. anonymous

Geezus, I just reread my work and I made another mistake. Sorry D: So when you put 1+t back in for u you get (1+t)(ln(1+t) -1) from 0 to 5. So when you put these values in, you get: 6(ln6 -1) - 1(ln1 -1) Which gives you the distance traveled. BUT you still have to add 2 to the answer, which will give you the position. So the answer should be B.

15. anonymous

i'm confused on the -1ln (1-1) part. i'm sorry lol

16. anonymous