# Thread: Reverse the order of integration

1. ## Reverse the order of integration

How do you do reverse order of integration for an iterated integral?

2. Originally Posted by ottmar0
How do you do reverse order of integration for an iterated integral?
You've asked what's called a fat question.

Can you give us the question you're working on so that we can guide you through it?

3. ## Here is the problem

double integral of the function of where y is greater than equal to tanpie/4 *x or less than or equal to x, and x is greater than 0 and less than or equal to 1.

4. Originally Posted by ottmar0
double integral of the function of where y is greater than equal to tanpie/4 *x or less than or equal to x, and x is greater than 0 and less than or equal to 1.
Is this right?

$\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{\int_0^1{f(x, y)\,dx}\,dy}$

5. ## change

switch them around, 1 with the x, and 0 with tan(piex/4) dydx.

6. Originally Posted by ottmar0
switch them around, 1 with the x, and 0 with tan(piex/4) dydx.
That's not what you asked for, but ok...

$\int_0^1{\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{f(x, y)\,dy}\,dx}$

How's this?

7. ## right.

thats it

8. Originally Posted by Prove It
Is this right?

$\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{\int_0^1{f(x, y)\,dx}\,dy}$
I find the easiest way to reverse orders of integration is to look at the inequalities and "solve" them simultaneously.

Notice that

$\tan{\left(\frac{\pi x}{4}\right)} \leq y \leq x$ and $0 \leq x \leq 1$.

You should be able to see that since $y \leq x$ and $x \leq 1$, then $y \leq 1$.

Also, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$ and $0 \leq x$, then $0 \leq y$.

Finally, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$, then

$\frac{\pi x}{4} \leq \arctan{y}$

$x \leq \frac{4}{\pi}\arctan{y}$.

$0 \leq y \leq 1$ and $y \leq x \leq \frac{4}{\pi}\arctan{y}$.

Thus the double integral is

$\int_0^1{\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{f(x, y)\,dy}\,dx} = \int_0^1{\int_y^{\frac{4}{\pi}\arctan{y}}{f(x, y)\,dx}\,dy}$.

9. Originally Posted by Prove It
I find the easiest way to reverse orders of integration is to look at the inequalities and "solve" them simultaneously.

Notice that

$\tan{\left(\frac{\pi x}{4}\right)} \leq y \leq x$ and $0 \leq x \leq 1$.

You should be able to see that since $y \leq x$ and $x \leq 1$, then $y \leq 1$.

Also, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$ and $0 \leq x$, then $0 \leq y$.

Finally, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$, then

$\frac{\pi x}{4} \leq \arctan{y}$

$x \leq \frac{4}{\pi}\arctan{y}$.

$0 \leq y \leq 1$ and $y \leq x \leq \frac{4}{\pi}\arctan{y}$.

Thus the double integral is

$\int_0^1{\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{f(x, y)\,dy}\,dx} = \int_0^1{\int_y^{\frac{4}{\pi}\arctan{y}}{f(x, y)\,dx}\,dy}$.
cool~! i've learned it in Calculus III
the changes of interval can also be explain graphically.. it would be the easiest explaination..

10. Originally Posted by Prove It
I find the easiest way to reverse orders of integration is to look at the inequalities and "solve" them simultaneously.

Notice that

$\tan{\left(\frac{\pi x}{4}\right)} \leq y \leq x$ and $0 \leq x \leq 1$.

You should be able to see that since $y \leq x$ and $x \leq 1$, then $y \leq 1$.

Also, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$ and $0 \leq x$, then $0 \leq y$.

Finally, if $\tan{\left(\frac{\pi x}{4}\right)} \leq y$, then

$\frac{\pi x}{4} \leq \arctan{y}$

$x \leq \frac{4}{\pi}\arctan{y}$.

$0 \leq y \leq 1$ and $y \leq x \leq \frac{4}{\pi}\arctan{y}$.

Thus the double integral is

$\int_0^1{\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{f(x, y)\,dy}\,dx} = \int_0^1{\int_y^{\frac{4}{\pi}\arctan{y}}{f(x, y)\,dx}\,dy}$.
thanks alot, its really helpful. I hate the graph sketching method and you just provided a simpler alternative. cheers