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

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Answer 1 · 2016-01-13T17:44:13

$f (x) = 4 x^{2} - x + 12$ $f(x)=4x^2-x+12$

By using normal rules of integration we can evaluate it as

$f (x) = \int (8 x - 1) d x$ $f(x)=int(8x-1)dx$

$= 4 x^{2} - x + C$ $=4x^2-x+C$ .

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

$f (0) = 12 \Leftrightarrow C = 12$ $f(0)=12iffC=12$

Therefore the required function is

$f (x) = 4 x^{2} - x + 12$ $f(x)=4x^2-x+12$ .

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