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

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Solution

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

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