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Find area between the curve y = 1/(1+x^2) and the x-axis. This is Calculus 2. I think I have a and b correct, but not sure of c. Any help would be very much appreciated! :)

(a) from -10 to 10 Answer that I got: arctan(10)-arctan(-10) = 2.942 (b) from -A to A Answer that I got: arctan(A)-arctan(-A) (c) Find the area under the whole curve. (Calculate the limit of your answer in part (b) as A → ∞) Have no idea, I think as A → ∞ then arctan(A) = pi/2, but do we also do it for -A? So would the answer be arctan(pi/2)-arctan(-pi/2)= 2?

Thanks for the help!

Temanibble
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1 Answers1

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Good job.

(a) is correct.

(b) is correct and can also be written 2*arctan(A),
because arctan(-A) = -arctan(A).

(c) careful here: arctan(A)→pi/2 as A → ∞, and arctan(-A) → -pi/2 as A → ∞,
so arctan(A) - arctan(-A) → pi.

Rich L
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