if a function F is such that F''(x) = sec⁴ x + 4

F'(0) =0 and F(0) = 0

find F(x)

just integrate twice the fiven f''(x). use condns. given b/w.

Integrate twice...

F'(x)=tanx+/3 +4x+c ...'c' comes out to be 0.

F(x)=/6+2-2log(cosx)/3

I may have integrated wrongly...coz i have done it mentally...but main thing is to integrate twice..thats it..

F'(x)=4x+sec²x(1+tan²x)dx=4x+tanx+1/3tan³x+c where c=0 as F'(0)=0

F(x)=2x²+tanx+1/3tanx(sec²x-1)dx=2x²+2/3lnIsecxI+1/6tan²x+c' where c'=0 as F(0)=0