What is #f(x) = int e^(3x)-e^(2x)+xdx# if #f(0)=-2 #?

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Answer 1 · 2018-03-04T22:40:28

#f(x)=1/3e^(3x)+1/2 (x^2-e^(2x)) -11/6#

#f(x) = int (e^(3x)-e^(2x)+x)dx#

#f(x)= inte^(3x)dx - inte^(2x)dx + intxdx#

#f(x)=1/3e^(3x)-1/2e^(2x)+1/2x^2 + C# ; where C#\equiv#constant

#f(x)=1/3e^(3x)+1/2 (x^2-e^(2x)) + C#

#f(0) = 1/3e^(3(0))-1/2e^(2(0))+cancel(1/2(0)^2) + C = -2#

#1/3e^(0)-1/2e^(0)+ C = -2#

#1/3(1)-1/2(1)+ C = -2#

#1/3 - 1/2 + C = -2#

#2/6 - 3/6 + C = -12/6#

# - 1/6 + C = -12/6#

#C = -11/6#

Hence:
#f(x)=1/3e^(3x)+1/2 (x^2-e^(2x)) -11/6#