Question

can't find the integral of this function
v(t) = (1+t^2)^(1/3)
anyone know how?

Answer 1

\[\int\limits \sqrt[3]{1 + t^2}\]

this is what i cant find

Answer 2

\[\large \int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\]

Answer 3

yep thats right

Answer 4

well, seems as though it doesn't have closed form based on my first impression.
Lets set u=1+t² and see wat we get, did you try that?

Answer 5

yep, didn't get anywhere. be my guest though im pretty bad at u substitutions :P

Answer 6

oh, my trig sub...

Answer 7

\(\large \displaystyle\int\limits_{ }^{ }\sqrt[3]{1+t^2}dt\)

\(\large \displaystyle t=\tan\theta \)

Answer 8

trig sub...? never heard of it.

Answer 9

lets see what then,
u=tan(theta)
du=sec^2theta
\(\large \displaystyle\int\limits_{ }^{ }\sec^2\theta\sqrt[3]{1+\tan^2\theta}~d\theta \)
\(\large \displaystyle\int\limits_{ }^{ }\sec^{2+2/3}\theta\)
that is bullsh... as well

Answer 10

wolfram is giving me something i dont even understand .

Answer 11

acc. to wolfram
http://www.wolframalpha.com/input/?i=integral+of+%281%2Bx%5E2%29%5E%281%2F3%29

Answer 12

no simple function answer

Answer 13

are you sure you aren't given limits of integration, and this is not a riemman sum type of question?

Answer 14

or maybe it is sort of

\(\large \displaystyle f(x)=\int\limits_{2 }^{ \cos(x)}\sqrt[3]{1+t^2}dt\)
and you need f'(x) ?

Answer 15

this is my given question

Answer 16

this is much better

Answer 17

can I use y instead of x
(as y(t) ?)

Answer 18

You can use a stepsize on that one I think

Answer 19

interesting i figured i just had to find s(t) and just change the constant and plug in t = 3

Answer 20

well, that is the task as far as the steps go, but s(t)=?

Answer 21

that is the question where we don't arrive at a simple conclusion just like this, right?

Answer 22

have you heard of a step-size, Euler's method or anything like this?

Answer 23

no sorry :(

Answer 24

\(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\)

y(0)=2, find y(3)=?
lets use a step-size of h=1 for now (but with h=1/2 or even less you can get a better approximation.

Each step of size h is as follows:

\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\)

Answer 25

maybe we won't understand why it is like this right now, but you can see the formula, without completely going like wt(f)....

Answer 26

\(\large \displaystyle \left(t_n,y_n\right)=\left(t_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(t_{n-1},y_{n-1})}~\right)\)

excuse me it should be t.

Answer 27

\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\)

you are given:
\(\large \displaystyle f'(t)=\sqrt[3]{1+t^2}\)
and a first point of (t=0, y=2)

Answer 28

\(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\)

this would be the first step starting from (0,2), and h=1

Answer 29

are you getting it at least a little? (sorry I am bad at explaining)

Answer 30

ok, I will give you the entire process wth h=1, and then you will tell me how much you get out of it.

Answer 31

lol sort of , i'm not quite sure what the h represents

Answer 32

you are starting from (0,2) right?
that is the first given point

so it is the size of the step, in other words, we are carefully drawing the approximation for the graph of y(t) based on y'(t) taking it with little stpes of size h (of size 1 in this case)

Answer 33

we don't know where this step size of 1 would land, and usiong this formula we can find that out.

Answer 34

and from there, when we do the step and find the next point, we will take another step... ultimately, we wil get to the y(3)

Answer 35

\(\large \displaystyle \left(x_n,y_n\right)=\left(x_{n-1}+h,~y_{n-1}~+h\cdot \color{red}{f'(x_{n-1},y_{n-1})}~\right)\)

\(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{f'(0,2)}~\right)\)

(there just isn't a y in this case, so would just plug in 0 for t)
I haven't ever done Euler's for explicitly defined functions.
\(\large \displaystyle \left(x_1,y_1\right)=\left(x_{0}+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\)

Answer 36

oh, you know x_0 is 0 , i forgot to put that in

Answer 37

okay wait one sec , why are we even finding (x1,y1)?

Answer 38

\(\large \displaystyle \left(x_1,y_1\right)=\left(0+1,~2~+1\cdot \color{red}{\sqrt[3]{1+0^2}}~\right)\)
\(\large \displaystyle \left(x_1,y_1\right)=\left(1,~3\right)\)

Answer 39

because that is the next point