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March 27, 2015

March 27, 2015

Posted by **Isaac** on Friday, July 12, 2013 at 7:54pm.

So I've drawn the enclosed region. To find the area would I use the Disc/shell method? If so the formula that I came up with looks like this:

If area = pi(r)^2 then it should be this...

= integral from 0 to 1 pi(arctan x)^2dx

is this correct? I took pi out to the front and then left arctanx^2.. what's the antiderivative of that? Im confused

- Calculus -
**Steve**, Saturday, July 13, 2013 at 4:27amIf you want area, discs and shells don't even come into play. You only use those if you want to revolve an area around an axis.

In this case, it's a straightforward integral:

a = ∫[0,1] arctan(x) dx

= x arctan(x) - 1/2 ln(x^2+1) [0,1]

= 1/4 (π - ln(4))

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