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

The question is find the area of the reagion that is bounded by the curve y=arctan x, x=0, x=1, and the x-axis.

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 4:27am
If 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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