Question

Help me to solve integration for [sqrt(2x" + x' + 2 dx)] .......

Answer 1

What are you trying to integrate? As it is, the problem you've written down doesn't make sense.

Answer 2

Lets say for example Ingration of ; \[\sqrt{2x^2 + 2 \sin 2x}\]

Answer 3

For that particular function there is no closed form for its integral in terms of elementary functions.

Answer 4

Sorry , i didnt get you...

Answer 5

Let me explain with the equation - attached ; Assume the x , y and z value .

Answer 6

That integral measures the length of a path. There is no general solution for that integral beyond the form in which you've given it.

But if you have a particular path, \( (x(t),y(t),z(t)) \), then you can try and evaluate it.

Answer 7

So , you mean that after I find those x' , y' and z' values , I just need to subsitute the values into the equations right . Then , how about the 3 - 0 integration . Do I just need to leave it ?

Answer 8

If you have a particular path, you find the integrand, \( \sqrt{ x'^2 + y'^2 + z'^2 }\) and then integrate it from 0 to 3, just as you would any other integral.

So the question is: what is the path whose length the question is you to find.

Answer 9

Mr.James , I would very much appreciate your help if you could help me to solve this .

Find the length of the curve
r(t)=<2t^3/2 , cos(2t) , sin(2t)>,
0<= t <= 3

Answer 10

You tell me for that path what is \( \sqrt{x'^2 + y'^2 + z'^2} \)

Answer 11

\[\sqrt{3t^{1/2} - 2 \sin 2t + 2 \cos 2t}\] - If this you are asking.....

Answer 12

I forgot the ^2 ....

Answer 13

Yes you did. Try again.

Answer 14

\[\sqrt{3t - 2 \sin^{2} 2t + 2 \cos^{2} 2t}\] .... Is it okay now....

Answer 15

ooopppsss....mistake again ...w8..

Answer 16

\[\sqrt{9t - 4 \sin ^{2} 2t + 4 \cos ^{2} 2t}\] .... Check plz....

Answer 17

Still wrong. Check your signs

Answer 18

Which signs your telling....Is it positive n negative issue....

Answer 19

\( y'^2 = ... \)

Answer 20

OKay w8.... Actually ; y = cos 2t.....so , y' = - 2 sin 2t....correct right....??

Answer 21

Hence \( y'^2 = +4 \sin^2 2t \)

Answer 22

But , the negative won't involved right...??

Answer 23

I can't stay and help you with this for ever. Show that that
\[ \sqrt{x'^2 + y'^2 + z'^2} = \sqrt{9t + 4} \]

Answer 24

Sorry for the trouble . Is \[\sqrt{9t + 4}\] the final answer . How about the \[\int\limits_{0}^{3} \sqrt{9t + 4}\] .... Please help me for the final time..

Answer 25

Repost your question on the left. Someone else should have time to chime in.

Answer 26

w8... i think i now the solution ; is it should be \[\sqrt{9t} + \sqrt{4} = 9t ^{1/2} + 2 =18t ^{3/2} + 2t .....\] Is it correct...

Answer 27

Absolutely not.

It is NOT the case that \( \sqrt{a+b} = \sqrt{a} + \sqrt{b} \).

It is likewise NOT the case that \( \sqrt{9t + 4} \ = \sqrt{9t} + \sqrt{4} \).

Answer 28

Will it be \[1 / 9 \int\limits_{0}^{3} ( 18t + 8 ) ^{3/2}\]

Answer 29

I mean \[1 / 9 [ (18t + 8)^{3/2}]_{0}^{3}\]