What is #f(x) = int (x+3)^2-2x dx# if #f(1) = 0 #?

1 Answer
Øko
Dec 19, 2017

#f(x)=1/3(x+3)^3-x^2-61/3#

Explanation:

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

Use #inth(x)+g(x)dx=inth(x)dx+intg(x)dx#
#f(x)=int(x+3)^2dx-int2xdx#
#f(x)=1/3(x+3)^3-x^2+C#

Evaluate the constant of integration
#0=1/3(1+3)^3-1^2+C#
#0=1/3(4)^3-1+C#
#0=64/3-3/3+C#
#0=61/3+C#
#C=-61/3#

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