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Calculate $\int_0^1\int_x^1$ $1\over 1+x^2$ dy dx

According to the book, integrating this directly would not be practical because I would have to use arctan and a calculator. It reverses the order of integration and switches the limits instead.

$\int_0^1\int_0^y$ $1\over 1+x^2$ dx dy

When I tried solving this, I ignored the function and just looked at the limits of integration. I graphed y=x and then shaded in the upper region from y=(0,1). Then I changed the equation from y=x to x=y and then moved the limits like that. Is this the correct way to do it?

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  • $\begingroup$ Is the method I used to find the answer the proper way to do it? $\endgroup$ – Elie Fraser Jun 3 '15 at 2:32
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    $\begingroup$ It is correct. Congrats ! $\endgroup$ – DeepSea Jun 3 '15 at 2:33
  • $\begingroup$ Thanks a ton NotALoner! =) $\endgroup$ – Elie Fraser Jun 3 '15 at 2:33
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Observe that one may evaluate your integral directly starting from the variable $y$: $$ \begin{align} \int_0^1\int_x^1 \frac1{1+x^2} dy \:dx&=\int_0^1 \frac1{1+x^2}\int_x^1 dy \:dx\\\\ &=\int_0^1 \frac1{1+x^2}(1-x) \:dx\\\\ &=\int_0^1 \frac1{1+x^2} \:dx-\frac12\int_0^1 \frac{2x}{1+x^2} \:dx\\\\ &=\frac{\pi}4-\frac{\ln2}2. \end{align} $$

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