Find F (x) from the given F'(x)
F'(x) = $3 x^{2}$ - 4x and F(0) = 1.

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Solution

$F^{'} (x) = 3 x^{2} - 4 x$
$\int F^{'} (x) d x = x^{3} - 2 x^{2} + c = F (x)$
Given F(0)=1
$F (0) = 0^{3} - {2.0}^{3} + c = 1 \Rightarrow c = 1$
$F (x) = x^{3} - 2 x^{2} + 1$

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