What is #f(x) = int (x+3)^2-3x dx# if #f(-1)=6 #?

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Answer 1 · 2015-12-31T06:43:20

#f(x) = x^3/3 +3/2x^2 +9x + 83/6#

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

#f(-1)# is given to find the constant of integration and we shall use it after the integration.

#int (x+3)^2 - 3x dx#

To simplify things we can expand #(x+3)^2#

#(x+3)^2 = x^2+6x+9#

#int (x+3)^2 - 3x dx #
#=int x^2 + 6x + 9 - 3x dx#
#=int x^2 +3x + 9 dx#
#= x^(2+1)/(2+1) + 3x^(1+1)/(1+1) + 9x + C#
#=x^3/3 - 3x^2/2 + 9x + C#

This is #f(x) = x^3/3 + 3x^2/2 + 9x + C#

We are given #f(-1) = 6# Let us use this

#f(-1) = (-1)^3/3 +3(-1)^2/2 + 9(-1) + C#
# 6 = -1/3+3/2 - 9 + C#
#6 = -2/6+9/6-54/6 + C#
#6=-47/6+C#
#6+47/6=C#
#36/6 + 47/6 = C#
#83/6 = C#

#f(x) = x^3/3 +3/2x^2 +9x + 83/6#