Evaluate integral from -1 to 2 of (2+x-x²) dx. Find the equation of the normal to the curve y=x+4/x at the point where x=4.

$$\left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} = \left( 2(2) + \frac{2^2}{2} - \frac{2^3}{3} \right) - \left( 2(1) + \frac{1^2}{2} - \frac{1^3}{3} \right)$$ $$= \left( 4 + \frac{4}{2} - \frac{8}{3} \right) - \left( 2 + \frac{1}{2} - \frac{1}{3} \right)$$ $$= \left( 4 + 2 - \frac{8}{3} \right) - \left( \frac{12}{6} + \frac{3}{6} - \frac{2}{6} \right)$$ $$= \left( 6 - \frac{8}{3} \right) - \left( \frac{13}{6} \right)$$ $$= \left( \frac{18}{3} - \frac{8}{3} \right) - \frac{13}{6}$$ $$= \frac{10}{3} - \frac{13}{6}$$

$$= \frac{20}{6} - \frac{13}{6} = \frac{7}{6}$$

The value of the integral is $\boxed{\frac{7}{6}}$.

: Equation of the Normal to a Curve

Find the equation of the normal to the curve $y = x + \frac{4}{x}$ at the point where $x=4$.

Substitute $x=4$ into the equation of the curve:

$$y = 4 + \frac{4}{4} = 4 + 1 = 5$$

The point is $(4, 5)$.

Rewrite $y = x + 4x^{-1}$.

$$\frac{dy}{dx} = 1 - 4x^{-2} = 1 - \frac{4}{x^2}$$

The gradient of the normal ($m_n$) is the negative reciprocal of the gradient of the tangent.

$$m_n = -\frac{1}{m_t} = -\frac{1}{3/4} = -\frac{4}{3}$$

Using point $(4, 5)$ and $m_n = -\frac{4}{3}$:

$$y - 5 = -\frac{4}{3}(x - 4)$$

Multiply by 3:

$$3(y - 5) = -4(x - 4)$$ $$3y - 15 = -4x + 16$$

Rearrange into the form $Ax + By = C$:

$$4x + 3y = 16 + 15$$ $$4x + 3y = 31$$

The equation of the normal is $\boxed{4x + 3y = 31}$.