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

The Correct Option is D

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:

\[ \boxed{\frac{4}{3}} \]