Question

The area enclosed between the curves $y = x|x|$ and $y = x - |x|$ is:

• A. $\frac{8}{3}$
• B. $\frac{2}{3}$
• C. 1
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

Answer 2

We calculate the area enclosed by the curves $y = x|x|$ and $y = x - |x|$ by analyzing their forms and intersections.

Step 1: Rewrite the curves

1. For $y = x|x|$:

y = \begin{cases} x^2 & \text{if } x \geq 0, \\\ -x^2 & \text{if } x < 0\. \end{cases}

This represents a parabola opening upwards in the first quadrant and downwards in the second quadrant.

2. For $y = x - |x|$:

y = \begin{cases} 0 & \text{if } x \geq 0, \\\ 2x & \text{if } x < 0\. \end{cases}

This represents the x-axis for $x \geq 0$ and a line with slope 2 for $x < 0$.

Step 2: Intersection points

In the second quadrant ($x < 0$), solve for the intersection points by setting:

$x|x| = x - |x| \implies -x^2 = 2x \quad (\text{as } x < 0, \text{ so } |x| = -x).$

Factorize:

$x(x + 2) = 0 \implies x = 0 \text{ or } x = -2.$

Thus, the curves intersect at $x = -2$ and $x = 0$.

Step 3: Define the enclosed area

The enclosed area is the integral of the difference between the upper curve $y = -x^2$ and the lower curve $y = 2x$, over the interval $x \in [-2, 0]$:

$A = \int_{-2}^0 \left[(-x^2) - (2x)\right] dx.$

Simplify:

$A = \int_{-2}^0 \left(-x^2 - 2x\right) dx.$

Step 4: Compute the integral

Evaluate term by term:

$\int -x^2 dx = -\frac{x^3}{3}, \quad \int -2x dx = -x^2.$

Thus:

$A = \left[ -\frac{x^3}{3} \right]_{-2}^0 + \left[ -x^2 \right]_{-2}^0.$

Step 5: Evaluate at bounds

1. For $\left[-\frac{x^3}{3}\right]_{-2}^0$:

$\left(-\frac{0^3}{3}\right) - \left(-\frac{(-2)^3}{3}\right) = 0 + \frac{-8}{3} = \frac{8}{3}.$

2. For $\left[-x^2\right]_{-2}^0$:

$-(0^2) + (-(-2)^2) = 0 - 4 = -4.$

Combine these:

$A = \frac{8}{3} - 4.$

Simplify:

$A = \frac{8}{3} - \frac{12}{3} = \frac{4}{3}.$

Final Answer: The enclosed area is:

$\boxed{\frac{4}{3}}$