Area integral

I'm doing 5b here.

\displaystyle\iint_R x^2 e^{-x^2}\,dx\,dy

R is the region inside the various curves on this plot, bounded by y = x, y = 0, y = 1/x and x = 2.

I split the region into the left and right parts (i.e < 1 and >= 1) but then I get (for example)

\displaystyle \int_0^x\int_0^1 x^2 e^{-x^2}\,dx\,dy

, the innermost part of which looks fairly hideous to solve (if at least one of the limits was

\pm\infty

, I could move on, but I don't know what to do here).

Thanks :smile:

Reply 1

generalebriety

16

I haven't looked at the question. However, your integral doesn't make sense. Your y-limits (0 and x) are dependent on x, which you're hoping to integrate out of the thing, so you should really be integrating w.r.t. y first. And, of course,

\displaystyle \int_0^1\int_0^x \left( x^2 e^{-x^2}\right)\,dy\,dx = \displaystyle \int_0^1 \left(x^2 e^{-x^2}\right)\cdot x\,dx

because

\int_0^x \,dy = x

.

Reply 2

Slumpy

16

generalebriety

I haven't looked at the question. However, your integral doesn't make sense. Your y-limits (0 and x) are dependent on x, which you're hoping to integrate out of the thing, so you should really be integrating w.r.t. y first. And, of course,

\displaystyle \int_0^1\int_0^x \left( x^2 e^{-x^2}\right)\,dy\,dx = \displaystyle \int_0^1 \left(x^2 e^{-x^2}\right)\cdot x\,dx

because

\int_0^x \,dy = x

.

Oops, misread, sorry!

Reply 3

DFranklin

18

Your notation looks a little confused - when you say:

\int_0^x

, what is "x" supposed to be?

In general, the "clever bit" in these questions is that if you choose the order of integration correctly, you find you get an "extra power of x" from the inner integral, which leaves you with something you can integrate in the outer integral as well.

e.g.

\displaystyle \int_{x=0}^{x=1} \int_{y=0}^{y=x} 2e^{-x^2}\,dy \,dx

looks like it's going to be horrible, what with that exp(-x^2) term that we can't integrate. But integrating the inner integral (w.r.t. y) gives us:

\displaystyle \int_{x=0}^{x=1} \Big[2y e^{-x^2}\Big]_0^x \,dx = \int_{x=0}^{x=1} 2x e^{-x^2}\,dx

which we can integrate by recognition (or sub t=x^2).

Reply 4

TheUnbeliever

OP

16

generalebriety

Your y-limits (0 and x) are dependent on x, which you're hoping to integrate out of the thing, so you should really be integrating w.r.t. y first.

A-ha, thanks. I think I've got the right answer now, because it then links in to part (a) and all falls out easily enough.

DFranklin

Your notation looks a little confused - when you say:

\int_0^x

, what is "x" supposed to be?

I'm not really sure I understand, sorry. :o:

The region there is bounded above by y = x and below by y = 0, so I was using them as the limits for the integral. Does it need to be written

\int_{y=0}^{y=x}

or am I missing your point?

In general, the "clever bit" in these questions is that if you choose the order of integration correctly, you find you get an "extra power of x" from the inner integral, which leaves you with something you can integrate in the outer integral as well.

Thanks. I think I see this, now. That's quite nice...

Reply 5

DFranklin

18

TheUnbeliever

I'm not really sure I understand, sorry. :o:

The region there is bounded above by y = x and below by y = 0, so I was using them as the limits for the integral. Does it need to be written

\int_{y=0}^{y=x}

or am I missing your point?If that's what you mean, then you need to swap the two integrals around.

Conceptually, you define the limits from the outside in. In an integral like:

\displaystyle \int_a^b \underbrace{\int^c_d f(x,y) \,dy}_{\text{function of x}} \, dx

Then a and b should not depend on either x or y, but the whole inner integral can be thought of as a function of x and in particular c and d can depend on x.

Reply 6

TheUnbeliever

OP

16

Oh, I see! Thanks, that makes much more sense - I was just ordering them more or less at random before. I missed these lectures because I was ill, which I've been regretting bitterly. :s-smilie:

Area integral

Related discussions

Latest

Trending

Trending

Articles for you