double integral

[mcwang719]

Reverse the order of integration and evaulate.
\(\displaystyle \int_0^1 {\int_y^1 {{2 \over {x^2 + 1}}dxdy} }\)
So what i did was \(\displaystyle \int_y^1 {\int_0^1 {{2 \over {x^2 + 1}}dydx} }\)
Now i integrated with respect to y and here's where im stuck
\(\displaystyle \int\limits_y^1 {{2 \over {x^2 + 1}}dx}\)
I'm having trouble integrating that, am i on the right track? can someone help? thanks!!!

 

[royhaas]

Go back and draw the region of integration before you reverse the order.

 

[tkhunny]

It's a definite integral. If you end up with variables in the final expression, you did something wrong.

 

[soroban]

Hello, mcwang719!

royhaas has the best advice.

Reverse the order of integration and evaulate: \(\displaystyle \:\L\int_0^{\;\;\;1}\int_y^{\;\;\;1} \frac{2}{x^2 + 1}\,dx\,dy\)

So what i did was: \(\displaystyle \L\,\int_y^{\;\;\;1} \int_0^{\;\;\;1} \frac{2}{x^2 + 1}\,dy\,dx\;\;\) . . . no

You do NOT simply switch the limits.


The original limits are: \(\displaystyle \,\begin{array}{cc}y\,=\,1\\ y\,=\,0\end{array}\) and \(\displaystyle \begin{array}{ccc}x\,=\,1\\ . \\ x\,=\,y\end{array}\)

The region looks like this:

      |
     1+       *(1,1)
      |     /:|
      |   /:::|
      | /:::::|
    --+-------+--
      0       1

Reversing the limits, we see that \(\displaystyle y\) goes from \(\displaystyle y\,=\,0\,\) up to \(\displaystyle \,y\,=\,x\)

\(\displaystyle \;\;\)and that \(\displaystyle x\) goes from \(\displaystyle \,x\,=\,0\,\) to \(\displaystyle \,x\,=\,1\)

The new integral is: \(\displaystyle \L\:\int^{\;\;\;1}_0 \int^{\;\;\;x}_0 \frac{2}{x^2\,+\,1}\,dy\,dx\)