integration question +r

Hmm...

Integrating from (0,0) to (1,1). We can use the parameter

x=y=t

,

t \in [0,1]

This gives the integral,

\displaystyle \int_0^1 2t \cdot 1 \, dt - \int_0^1 t^4 \cdot 1 \, dt = \int_0^1 (2t-t^4) \, dt = \boxed{\frac{4}{5}}

I'm not sure exactly what you've done?

Reply 2

OP

Ah yeah for that bit I did the same method but had a sign error, now I get 4/5.

How about the rest?

Reply 3

I disagree with your sign for the second one.

We parametrise the path from (1,1) to (0,1) as x=1-t, y=1,

t \in [0,1]

Therefore we get the integral

\displaystyle \int_0^1 2 \cdot 1 \cdot ( 0 ) \, dt - \int_0^1 1 \cdot (1-t)^3 \cdot (-1) \, dt = \int_0^1 (1-t)^3 \, dt = \frac{1}{4}

?

I also disagree with your sign on the last part

Reply 4

I am of course, assuming the direction is (0,0)->(1,1)->(0,1)

Reply 5

OP

SimonM

I disagree with your sign for the second one.

We parametrise the path from (1,1) to (0,1) as x=1-t, y=1,

t \in [0,1]

Therefore we get the integral

\displaystyle \int_0^1 2 \cdot 1 \cdot ( 0 ) \, dt - \int_0^1 1 \cdot (1-t)^3 \cdot (-1) \, dt = \int_0^1 (1-t)^3 \, dt = \frac{1}{4}

?

I also disagree with your sign on the last part

Hmm, the way I did it was [for the second part [and yes the direction is correct]]:
y=1, 0<x<1 so dy=0 [no change in y] so

\int 2ydy - yx^3dx

goes to

\int_0^1 2dy - x^3 dx = \int_0^1-x^3 dx =-\frac{1}{4}

I'm also following methods my lecturer used just to let you know =) Thanks for the help so far btw.

Reply 6

Shadow!

I'm also following methods my lecturer used just to let you know =) Thanks for the help so far btw.

Remember, the "direction" of x is ranging from 1 to 0, rather than 0 to 1

Reply 7

OP

SimonM

Remember, the "direction" of x is ranging from 1 to 0, rather than 0 to 1

So if the direction goes from 1 to 0 instead of 0 to 1, then we integrate with the integrating values switched? So

\int_0^1 F(x) dx = -\int_1^0 F(x) dx

which would be why we had opposite signs...

Reply 8

Yes, exactly.

That said, your professor might prefer you to say something about dx. I don't know how he's taught you

Reply 9

OP

SimonM

Yes, exactly.

That said, your professor might prefer you to say something about dx. I don't know how he's taught you

I think you may be right about that. That being said, is what I have done for the third and last part of the curve correct? Is it right to integrate with respect to y after doing the first two parts with respect to x?

Reply 10