$\frac{d y}{d x} = 3 x^{2} + 2 x + 1$ find $y$ when $x = 1$

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Solution

$\frac{d y}{d x} = 3 x^{2} + 2 x + 1 y = \int 3 x^{2} + 2 x + 1 d x y = \frac{3 x^{3}}{3} + \frac{2 x^{2}}{2} + x y = x^{3} + x^{2} + x y = 1^{3} + 1^{2} + 1 = 3$

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