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anonymous
 4 years ago
help with calclulus
anonymous
 4 years ago
help with calclulus

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can 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
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0LOL but why didthey do that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry abt the attachment but there seems to be a bug

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0its to make the numerator look like the derivative of the denominator.... which would suggest the solution would be a ln function

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0the 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
 4 years ago
Best ResponseYou've already chosen the best response.0oops should read 1/2 tan^(1) (u) then re substitute 1/2 tan^(1) (y^2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks that was clear campbell :D That is exactly what I wanted to know
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