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

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Answer 1 · 2016-11-19T23:37:08

Simply expand #(x- 1)^2# and then integrate using the rule #intx^n(dx) = x^(n + 1)/(n + 1) + C#.

#f(x) = intx^2 - 2x + 1 + 3xdx#

#f(x) = intx^2 - x + 1dx#

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

We can solve for #C# now.

When #x = 3#, #y = 2#:

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

#2 = 18 - 9/2 + 3 + C#

#-19 + 9/2= C#

#-29/2 = C#

Hence, #f(x) = 2/3x^3 - 1/2x^2 + x - 29/2#.

Hopefully this helps!