$y = x^2 + 2x$
$x^2 + 2x + 1 = y + 1$
$(x + 1)^2 = y + 1$ → upward parabola, vertex at (-1, -1), LR = 1
$y = x^3 - 4x$
$y' = 3x^2 - 4 = 0$
$x^2 = \frac{4}{3}$
$x = \pm 1.15$ → (approximate - for graphing purposes only)
Maximum and minimum points of y = x3 - 4x
when x = 1.15, y = 1.153 - 4(1.15) = -3.08
when x = -1.15, y = (-1.15)3 - 4(-1.15) = 3.08
Maxima: (-1.15, 3.08)
Minima: (1.15, -3.08)
Points of intersection:
$y_{cubic} = y_{parabola}$
$x^3 - 4x = x^2 + 2x$
$x^3 - x^2 - 6x = 0$
$x(x^2 - x - 6) = 0$
$x(x - 3)(x + 2) = 0$
$x = 0, \, 3, \, \text{ and } \, -2$
when x = 0, y = 0
when x = 3, y = 3^3 - 4(3) = 15
when x = -2, y = (-2)^3 - 4(-2) y = 0
Points of intersection: (0, 0), (3, 15) and (-2, 0)
$A_1 = \displaystyle{{\int_{x_1}}^{x_2}} y_1 \, dx$
$A_1 = \displaystyle{{\int_{x_1}}^{x_2}} (y_U - y_L) \, dx$
$A_1 = \displaystyle{{\int_{-2}}^0} [ \, (x^3 - 4x) - (x^2 + 2x) \, ] \, dx$
$A_1 = \displaystyle{{\int_{-2}}^0} (x^3 - x^2 - 6x) \, dx$
$A_1 = \left[ \dfrac{x^4}{4} - \dfrac{x^3}{3} - 3x^2 \right]_{-2}^0$
$A_1 = \left[ \dfrac{0^4}{4} - \dfrac{0^3}{3} - 0^2 \right] - \left[ \dfrac{(-2)^4}{4} - \dfrac{(-2)^3}{3} - (-2)^2 \right]$
$A_1 = 0 - (-\frac{16}{3})$
$A_1 = \frac{16}{3} \, \text{ unit}^2$ answer
$A_2 = \displaystyle{{\int_{x_1}}^{x_2}} y_2 \, dx$
$A_2 = \displaystyle{{\int_{x_1}}^{x_2}} (y_U - y_L) \, dx$
$A_2 = \displaystyle{{\int_0}^3} [ \, (x^2 + 2x) - (x^3 - 4x) \, ] \, dx$
$A_2 = \displaystyle{{\int_0}^3} (6x + x^2 - x^3) \, dx$
$A_2 = \left[ 3x^2 + \dfrac{x^3}{3} - \dfrac{x^4}{4} \right]_0^3$
$A_2 = \left[ 3(3)^2 + \dfrac{3^3}{3} - \dfrac{3^4}{4} \right] - \left[ 3(0)^2 + \dfrac{0^3}{3} - \dfrac{0^4}{4} \right]$
$A_2 = \frac{63}{4} \, \text{ unit}^2$ answer
