Area question

[dannatyler]

Find the area under the graph of y= 1/1+x^2 on the interval (-1,1). Here is what I came up with:
x=1
INT. 1/1+x^2 dx = arctanx
x= -1

arctan(1)-arctan(-1)= pi/2

 

[galactus]

Yep. Good show. Keep this 'identity' in mind. You will see it later.

What I mean is knowing that the derivative of arctan is 1/(1+x^2) and the antiderivative of 1/(1+x^2) is arctan will arise later, and it is handy to make the observation.