Why Do I Get Different Answers When Changing the Order of Integration?

[seanc12]

Homework Statement

Solve the following Double Intergral, and show the answer is the same, regardless of which order you integrate.
The integral is between the boundaries y=x and y=x^2

Homework Equations

\int\int_R (x^2 + 2y)dxdy

The Attempt at a Solution

So first of all i integrated with repsect to y first

\int_0^1 dx \int^{x}_{x^2} (x^2 + 2y)dy

\int_0^1 dx [yx^2+y^2]^x^2_x

\int_0^1 dx (x^3 + x^2 - 2x^4)

which give me an answer of
\frac{11}{60}Then, with respect to x first I get:

\int_0^1 dy \int_{\sqrt{y}}^y dx (x^2 + 2y)

\int_0^1 dy [\frac{x^3}{3} + 2xy ]^{y}_{\sqrt{y}}

\int_0^1 dy(\frac{y^3}{3} + 2y^2 - \frac{\sqrt{y}^3}{3} - \frac{2\sqrt{y}y}{3})

This leads me to an answer different to what I got from differenciating with repect to y first.

Can someone please enlighten me with what I am doing wrong.

 

[Disconnected]

Howdy, Sean!
Well, it looks like you may have made an error concerning the limits of the integrals!

It seems to me that the limits for dy should be x^2 to x, and the limits for x should by 0 to 1.

When you said you would integrate W.R.T. y first, you got rid of the dx... So I am assuming that is a typo.

Another typo is at the square bracket, you have it as x to x^2 rather then x^2 to x.

But the main problem seems to be your rearranging of variables. Remember that we are looking at an area. Try drawing it. It should quickly become apparent that when doing x first, in terms of y, the SQRT(y) is greater then y. Thus it should go FROM y TO SQRT(y).

This will give you the correct solution.

 

[seanc12]

Hi, thanks for the reply

Yeah, sorry about all the typos, I'm new to LaTeX so its a bit hard to follow some of what I'm typing.

I actually had my integral with repect to y from x to x^2, it was a typo.

I keep getting myself confused with the limits I should take, but hopefully I have it figured out now.

Thanks for the help!

 

[Disconnected]

No problem. I have issues with the limits as well.

What I find works extremely well is to draw a graph. Then you can just simply look at the graph and see what is going on, keeping in mind that what the integral at each limit is giving is the area between that curve and the axis, so the area between the is the area under the curve farthest from the axis minus the area under the curve closest to the axis.