Question

Can anyone help me find the integral or int(0,2) 1/(1+x^2)?

I get that the anti-derivative of 1/(1+x^2) would be arctan(x), so would I just equate (arctan(2)-arctan(1))?

Answer 1

arctan(2) - arctan(0)

Answer 2

So thats all i have to do? My book wants me to use "property 8" which is: m less than equal to f(x) less than equal to M for a less than equal to x less than equal to b, then

m(b-a) less than equal to Int (from a to b) f(x)dx less than equal to M(b-a)

Answer 3

I am not sure what you are asking now

Answer 4

you want to approximate this integral

Answer 5

would the answer to arctan(2) - arctan(0) be my final answer?

Answer 6

the integral of 1/(1+x^2) on [0,2] is equal to arctan(2) - arctan(0).

If you want to approximate it using minimum and maximum
the area is between 2*f(2) <= area <= 2 * f(0)

Answer 7

2 * 1/( 1 + 2^2) <= integral {0,2} 1/(1+x^2) <= 2 * 1 / ( 1 + 0^2)