# Integral what has gone wrong here?

1. Feb 27, 2006

### mmh37

I am still struggling my way through line integrals, and this here is one where I do not understand what has gone wrong - does anyone see what it is ( I really want to understand all of this)?

$$\int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy }$$

the curve joins (0,0) to (1,1)

where x= t and y=t (0<= t <= 1)

so, I said that

$$\frac {dx} {dt} = \frac {dy} {dt} = 1$$

therefore:
$$\int_{0}^{1} {( t*e^{t^2} + 2t + t + t*e^t^2 + t)dt} = \int_{0}^{1}{ (2t*e^{t^2} + 4t) dt} = e^1 + 1$$

However, the solution is supposed to be -2pi

Last edited: Feb 28, 2006
2. Feb 27, 2006

### assyrian_77

What do you mean by $$e^{t2}$$? Is it $$e^{2t}$$ or $$e^{t^2}$$?

I guess it comes from the first expression, viz. $$ye^xy$$. Is it $$y^2e^x$$ or $$ye^{xy}$$?

Never mind, going through your calculation, I can see what you meant.

Hmmm... I obtain the same thing as you. Are you sure about that parametrization?

Last edited: Feb 27, 2006
3. Feb 28, 2006

### mmh37

I am so sorry about that.....I fixed it.

That's the parametrization the question stated:

(....) the curve joins (0,0) to (1,1)

where x= t and y=t (0<= t <= 1)

( I just double checked this)

Last edited: Feb 28, 2006
4. Feb 28, 2006

### mmh37

does it mabe help that the field is conservative?

Last edited: Feb 28, 2006
5. Feb 28, 2006

### HallsofIvy

Staff Emeritus
A bit. That of course is the reason the path did not need to be given!

And it verifies that the correct answer is e+ 1.