Question

can anyone help me in finding the area between two curves, I got wrong answer , (68/1+x^2) and y=|x|

Answer 1

34pi-4 is my answer

Answer 2

Very close, probably some minor mistake in the integration.
Can you post your integrand and limits of integration?

Answer 3

\[\int\limits_{0}^{1} \tan^{-1} (x)-x^2\]

Answer 4

I would like to see the integrand, i.e. before integration or any substitution.
Also, did you make a sketch of the functions, and determined the intersection points?

Answer 5

yes , 1 and -1

Answer 6

Well, I get limits of -4 and 4, which also shows by inspection that both functions are even. This means that we can integrate from 0 to 4 and double the value for the answer.

Answer 7

Apart from the limits, you may consider the integrand as
68/(1+x^2)-x.
What you gave me could be some part of the antiderivative.

Answer 8

so instead of a=0 b=1 , I must use a=0 b=4?

Answer 9

yes, something like:
\(2\int_0^4(\frac{68}{1+x^2}-x)dx\)

Answer 10

thanks @mathmate , last question , how did you get the -4 and 4 ? when I am solving it I always ger 1, -1 as points of intersection

Answer 11

thanks , I already got the answer , I clicked wrong number , thanks anyway

Answer 12

ok, you're welcome!

Answer 13

so the correct answer is 164.3112023? 136(tan^-1(4)-tan^-1(0))-1?

Answer 14

136(tan^-1(4)-tan^-1(0))-4^2? am I right @mathmate ?

Answer 15

Use \(tan^{-1}0 = 0\) to simplify your answer.
Also the integral of x is \(\frac{x^2}{2}\).
Do not forget to multiply by two for the final answer (136 is correct).