Integral problem reverse order

[Damascus Road]

Integral problem "reverse order"

Greetings,
This is from an online homework question:

Evaluate the integral by reversing the order of integration.

$$\int^{1}_{0}\int^{8}_{8y} e^{x^{2}}dxdy$$

Although, I fail to see how this works, if I switch the order I get:


$$\int^{8}_{8y}\int^{1}_{0} e^{x^{2}}dydx$$


after integrating wrt y, it boils down to

$$\int^{8}_{8y} e^{x^{2}}dy$$

which won't give me a number... did I do something wrong?

 

[Mute]

The limits of your integral depend on y, so you can't just switch the order of integration without changing the limits accordingly. Your integration region is {y: [0,1], x: [8y,8]}, so to switch the order of integration you need to figure out what the limits of y are in terms of x and what the constant limits of x are.

It helps to draw a picture of the region of integration. y runs from 0 to 1 along the y-axis, and since x is from 8y to 8, it runs from 0 to 8 along the x-axis, but you're only considering the area below the line x = 8y, or equivalently y = x/8.

So, the limits could equivalently be written as x: [0,8], y:[x/8,1]. So, when you go to integrate over y, you need to switch to these limits when you integrate.

 

[Damascus Road]

Ohhh, thanks Mute. My bad.

So, after integrating wrt y, it becomes

$$\int^{8}_{0}[e^{x^{2}} - \frac{x}{8}e^{x^{2}}] dx$$

requiring integration by parts, with a substitution...etc. Yes?

 

[HallsofIvy]

The outer integral is from 0 to 1/8, not from 0 to 8.
$$\int xe^{x^2} dx$$
can be integrated by the substitution u= x².

But
$\int e^{x^2} dx[/tex[<br /> does not have any elementary function as it anti-derivative.$

 

[Damascus Road]

So, your saying y runs from 0 to 1/8 and x runs from 1 to 1/8?

that doesn't make sense to me...