Integration

Term by term: $4x^3 \to \dfrac{4x^4}{4} = x^4$; $-6x \to \dfrac{-6x^2}{2} = -3x^2$; the constant $5 \to 5x$. Add $+c$.

$\left[3x^2\right]_1^2 = 3(4) - 3(1) = 12 - 3 = 9$.

$\left[x^3 - x^2\right]_0^2 = (8 - 4) - (0 - 0) = 4$.

Integrating gives $y = 2x^2 - 3x + c$. At $(1,2)$: $2 = 2 - 3 + c$, so $c = 3$.

The curve meets the axis where $4 - x^2 = 0$, so $x = -2$ and $x = 2$. Area $= \displaystyle\int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2} = \left(8 - \dfrac83\right) - \left(-8 + \dfrac83\right) = 16 - \dfrac{16}{3} = \dfrac{32}{3}$.

$v = \displaystyle\int (6t - 4)\,dt = 3t^2 - 4t + c$. Starting from rest means $v = 0$ at $t = 0$, so $c = 0$. Then $v(3) = 27 - 12 = 15$.