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))
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
1 answer
1 answer