What is #f(x) = int (x+3)^2+3x dx# if #f(5)=2 #?

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Answer 1 · 2017-05-20T16:25:11

#f(x) = 1/3 x^3 + 9/2 x^2 + 9x - 1183/6#

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

This integral could be solved by doing a #u# substitution where #u=x+3#, OR by expanding out the binomial in the parentheses.

I'll expand out the value in the parentheses for simplicity (and to use less extraneous variables).

#f(x) = int(x^2+6x+9+3x)dx#

#f(x) = int(x^2+9x+9)dx#

#f(x) = 1/3 x^3 + 9/2 x^2 + 9x +C#

Plug in the initial condition and solve for #C# (an arbitrary constant):
#f(5) = 2#

#2 = 1/3 (5)^3 + 9/2 (5)^2 + 9(5) +C#

I used a calculator in this step (not necessary)
#C = (-1183)/6 approx -197.1667#

Using this value for #C# in the equation for #f(x)#:

#f(x) = 1/3 x^3 + 9/2 x^2 + 9x - 1183/6#