What is #f(x) = int 8x-1 dx# if #f(0)=12 #?

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Answer 1 · 2016-01-13T17:44:13

#f(x)=4x^2-x+12#

By using normal rules of integration we can evaluate it as

#f(x)=int(8x-1)dx#

#=4x^2-x+C#.

But since the initial boundary condition was given, we may substitute it in to find

#f(0)=12iffC=12#

Therefore the required function is

#f(x)=4x^2-x+12#.