help with calclulus

Question

help with calclulus

anonymous · Answer

attachment

anonymous · Answer

can you explain to me what was done?
I undertsnad what they did with the infinity but I dont get how they added a 2 to the integrand

campbell_st · Answer

if you look at the intergrand they have put 1/2 out the front and the numerator is 2y...... remember 1/2 x 2y = y.....

anonymous · Answer

LOL but why didthey do that?

anonymous · Answer

sorry abt the attachment but there seems to be a bug

campbell_st · Answer

its to make the numerator look like the derivative of the denominator.... which would suggest the solution would be a ln function

campbell_st · Answer

the solution might be... let u = y^2   du = 2y

$$1/2\int\limits_{0}^{b}  du/(u^2 + 1)$$

which is tan^(-1)

campbell_st · Answer

oops should read 1/2 tan^(-1) (u)   then re substitute    1/2 tan^(-1) (y^2)

anonymous · Answer

Thanks that was clear campbell :D That is exactly what I wanted to know