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

The area bounded by the curves \(y = x|x|\) and \(y = x - |x|\) is determined by analyzing their representations and intersection points.

Step 1: Curve Representation

1. The curve \(y = x|x|\) is defined as:

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

This curve is a parabola opening upwards for \(x \geq 0\) and downwards for \(x<0\).

2. The curve \(y = x - |x|\) is defined as:

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

For \(x \geq 0\), this is the x-axis. For \(x<0\), this is a line with a slope of 2.

Step 2: Intersection Points

To find intersections in the region where \(x<0\), we set the equations equal:

\[ x|x| = x - |x| \]

Since \(x<0\), \(|x| = -x\). Substituting this gives:

\[ -x^2 = x - (-x) \implies -x^2 = 2x \]

Rearranging and factoring:

\[ x^2 + 2x = 0 \implies x(x + 2) = 0 \]

The solutions are \(x = 0\) and \(x = -2\). Therefore, the curves intersect at \(x = -2\) and \(x = 0\).

Step 3: Area Definition

The enclosed area is calculated by integrating the difference between the upper and lower curves over the interval \([-2, 0]\). In this interval, \(y = -x^2\) is the upper curve and \(y = 2x\) is the lower curve.

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

This simplifies to:

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

Step 4: Integral Computation

The antiderivatives of the terms are:

\[ \int -x^2 dx = -\frac{x^3}{3} \quad \text{and} \quad \int -2x dx = -x^2 \]

Applying these to the definite integral:

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

Step 5: Evaluation at Limits

1. Evaluating \(\left[-\frac{x^3}{3}\right]_{-2}^0\):

\[ \left(-\frac{0^3}{3}\right) - \left(-\frac{(-2)^3}{3}\right) = 0 - \left(-\frac{-8}{3}\right) = 0 - \frac{8}{3} = -\frac{8}{3} \]

2. Evaluating \(\left[-x^2\right]_{-2}^0\):

\[ -(0^2) - (-(-2)^2) = 0 - (-(4)) = 0 + 4 = 4 \]

Combining the results for the total area:

\[ A = -\frac{8}{3} + 4 \]

Simplifying:

\[ A = -\frac{8}{3} + \frac{12}{3} = \frac{4}{3} \]

Final Result: The calculated enclosed area is:

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