What is #f(x) = int x-x^2secx dx# if #f(pi/4)=3 #?

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Answer 1 · 2018-08-13T21:07:08

#int(x-x^2)dx=x^2/2-x^3/3+3+pi^3/192-pi^2/32# if #f(pi/4)=3#

#I=int(x-x^2)dx#

#=intxdx-intx^2dx#

#=x^2/2-x^3/3+C#, #C in RR#

Also, #f(pi/4)=3#

#(pi/4)^2/2-(pi/4)^3/3+C=3#

#C=3+pi^3/192-pi^2/32#

Finally, #I=x^2/2-x^3/3+3+pi^3/192-pi^2/32#

\0/ here's our answer !