Math Archive (14)

Given that $\begin{cases} y = x^2 + x - 2 &...(1) \\ y = 2x & ...(2) \end{cases}$

$\begin{aligned} (1)=(2): \quad x^2+x-2 & = 2x \\ x^2 - x - 2 & = 0 \\ (x+1)(x-2) & = 0 \end{aligned}$

Therefore, the bound area by $y = x^2 + x - 2$ and $y = 2x$ is between $-1 \le x \le 2$ . We need to find the range of $x$ that satisfies $|x^2+x-2|+|2x|=|x^2+3x-2|$ . Let us consider the $LHS$ of the equation.

$\begin{aligned} |x^2+x-2|+|2x| & = |(x+2)(x-1)|+|2x| \\ & = \begin{cases} x^2+x-2+2x = x^2+3x-2 = RHS & \text{for }x \ge 1 \\ -(x^2+x-2) + 2x = -x^2+x+2 \ne RHS & \text{for }0 \le x < 1 \\ -(x^2+x-2) - 2x = \color{#3D99F6}{- (x^2+3x-2) = RHS} & \text{for } -2 \le x < 0 \end{cases} \end{aligned}$

Note that for $-2 \le x < 0$ , $\color{#3D99F6}{x^2+3x-2} < 0 \implies \color{#3D99F6}{- (x^2+3x-2) = |x^2+3x-2| = RHS }$ .

Therefore, the ranges of $x$ that satisfy $|x^2+x-2|+|2x|=|x^2+3x-2|$ are $-1 \le x \le 0$ and $1 \le x \le 2$ and the area bounded by the ranges is:

$\begin{aligned} A & = \left| \int_{-1}^0 \left(x^2+x-2 - 2x \right) dx \right| + \left| \int_1^2 \left(x^2+x-2 - 2x \right) dx \right| \\ & = \left| \int_{-1}^0 \left(x^2-x-2 \right) dx \right| + \left| \int_1^2 \left(x^2-x-2 \right) dx \right| \\ & = \left| \frac{x^3}{3} - \frac{x^2}{2} - 2x \bigg|_{-1}^0 \right| + \left| \frac{x^3}{3} - \frac{x^2}{2} - 2x \bigg|_1^2 \right| \\ & = \left| 0 - \frac{7}{6} \right| + \left| -\frac{20}{6} + \frac{13}{6} \right| \\ & = \frac{7}{6} + \frac{7}{6} = \boxed{\dfrac{7}{3}} \end{aligned}$