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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 - ,

    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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