## anonymous 4 years ago help with calclulus

1. anonymous

2. anonymous

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

3. campbell_st

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.....

4. anonymous

LOL but why didthey do that?

5. anonymous

sorry abt the attachment but there seems to be a bug

6. campbell_st

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

7. campbell_st

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

8. campbell_st

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

9. anonymous

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