Finding the Area Bounded by Curves \(f(x) = x|x| - 1\) and \(g(x)\)

The problem asks us to find the area enclosed by two curves given by the functions \(f(x) = x|x| - 1\) and the piecewise function \(g(x) = \left\{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{\;x}} > 0}\\ {2{\rm{x}},{\rm{\;x}} \le 0} \end{array}} \right.\).

First, let's rewrite the function \(f(x)\) as a piecewise function based on the definition of the absolute value:

• If \(x \ge 0\), then \(|x| = x\). So, \(f(x) = x(x) - 1 = x^2 - 1\).
• If \(x < 0\), then \(|x| = -x\). So, \(f(x) = x(-x) - 1 = -x^2 - 1\).

Thus, \(f(x)\) can be written as: \({f}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{{\rm{x}}}^2 - 1,{\rm{\;x}} \ge 0}\\ {-{{\rm{x}}}^2 - 1,{\rm{\;x}} < 0} \end{array}} \right.\)

Now we have both functions defined piecewise:

Finding Intersection Points of the Curves

To find the area bounded by the curves, we need to determine where they intersect. We set \(f(x) = g(x)\) for different intervals.

Case 1: Intersection for \(x > 0\)

For \(x > 0\), \(f(x) = x^2 - 1\) and \(g(x) = \frac{3x}{2}\). Set them equal:

\(x^2 - 1 = \frac{3x}{2}\)

\(2x^2 - 2 = 3x\)

\(2x^2 - 3x - 2 = 0\)

We can factor this quadratic equation:

\((2x + 1)(x - 2) = 0\)

This gives potential solutions \(x = -\frac{1}{2}\) and \(x = 2\). Since we are considering the case \(x > 0\), the valid intersection point in this interval is \(x = 2\).

Case 2: Intersection for \(x \le 0\)

For \(x \le 0\), \(f(x) = -x^2 - 1\) and \(g(x) = 2x\). Set them equal:

\(-x^2 - 1 = 2x\)

\(0 = x^2 + 2x + 1\)

This is a perfect square trinomial:

\(0 = (x + 1)^2\)

This gives the solution \(x = -1\). Since we are considering the case \(x \le 0\), the valid intersection point in this interval is \(x = -1\).

The intersection points are at \(x = -1\) and \(x = 2\). These points define the limits of integration for the bounded area.

Determining the Upper and Lower Curves

The interval over which we need to calculate the area is \([-1, 2]\). Since both functions are piecewise defined with a split at \(x=0\), we need to consider the intervals \([-1, 0]\) and \([0, 2]\) separately.

Interval \([-1, 0]\):

In this interval, \(f(x) = -x^2 - 1\) and \(g(x) = 2x\). Let's pick a test point, say \(x = -0.5\).

• \(f(-0.5) = -(-0.5)^2 - 1 = -0.25 - 1 = -1.25\)
• \(g(-0.5) = 2(-0.5) = -1\)

Since \(-1 > -1.25\), \(g(x)\) is the upper curve and \(f(x)\) is the lower curve in the interval \([-1, 0]\).

Interval \([0, 2]\):

In this interval, \(f(x) = x^2 - 1\) and \(g(x) = \frac{3x}{2}\). Let's pick a test point, say \(x = 1\).

• \(f(1) = 1^2 - 1 = 0\)
• \(g(1) = \frac{3(1)}{2} = 1.5\)

Since \(1.5 > 0\), \(g(x)\) is the upper curve and \(f(x)\) is the lower curve in the interval \([0, 2]\).

In both intervals, \(g(x) \ge f(x)\).

Calculating the Area Bounded by the Curves

The total area is the sum of the areas over the two sub-intervals:

\(\text{Area} = \int_{-1}^{2} (g(x) - f(x)) dx = \int_{-1}^{0} (g(x) - f(x)) dx + \int_{0}^{2} (g(x) - f(x)) dx\)

Area over \([-1, 0]\):

\(\text{Area}_1 = \int_{-1}^{0} (2x - (-x^2 - 1)) dx = \int_{-1}^{0} (x^2 + 2x + 1) dx = \int_{-1}^{0} (x+1)^2 dx\)

Evaluate the integral:

\(\text{Area}_1 = \left[ \frac{(x+1)^3}{3} \right]_{-1}^{0}\)

\(\text{Area}_1 = \frac{(0+1)^3}{3} - \frac{(-1+1)^3}{3} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}\)

Area over \([0, 2]\):

\(\text{Area}_2 = \int_{0}^{2} (\frac{3x}{2} - (x^2 - 1)) dx = \int_{0}^{2} (-x^2 + \frac{3x}{2} + 1) dx\)

Evaluate the integral:

\(\text{Area}_2 = \left[ -\frac{x^3}{3} + \frac{3x^2}{4} + x \right]_{0}^{2}\)

\(\text{Area}_2 = \left( -\frac{2^3}{3} + \frac{3(2)^2}{4} + 2 \right) - \left( -\frac{0^3}{3} + \frac{3(0)^2}{4} + 0 \right)\)

\(\text{Area}_2 = \left( -\frac{8}{3} + \frac{12}{4} + 2 \right) - (0)\)

\(\text{Area}_2 = -\frac{8}{3} + 3 + 2 = -\frac{8}{3} + 5\)

\(\text{Area}_2 = \frac{-8 + 15}{3} = \frac{7}{3}\)

Total Area:

The total area bounded by the curves is the sum of the areas over the two intervals:

\(\text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{1}{3} + \frac{7}{3} = \frac{8}{3}\)

The area bounded by the curves is \(\frac{8}{3}\) square units.

Revision Table: Area Bounded by Curves

Additional Information on Bounded Area Calculation

Finding the area bounded by curves is a fundamental application of definite integrals in calculus. The general procedure involves several key steps:

• Graphing (Optional but Helpful): Sketching the curves can help visualize the region and identify the intersection points and which function is the upper one.
• Finding Intersection Points: Set the function equations equal to each other and solve for \(x\) (or \(y\) if integrating with respect to \(y\)). These points determine the limits of integration.
• Identifying the Upper and Lower Curves: In the interval between intersection points, determine which function has larger values. This function will be the "upper" curve, and the other will be the "lower" curve. If the curves cross within the interval, you'll need to split the interval at the crossing points and set up separate integrals. The difference \((\text{Upper Curve Function}) - (\text{Lower Curve Function})\) must be non-negative over the interval of integration.
• Setting up the Definite Integral: The area between two curves \(f(x)\) and \(g(x)\) from \(x=a\) to \(x=b\), where \(f(x) \ge g(x)\) over \([a, b]\), is given by \(\int_{a}^{b} (f(x) - g(x)) dx\). If the roles of upper and lower switch or if functions are defined piecewise, the integral must be split accordingly.
• Evaluating the Integral: Compute the definite integral using the Fundamental Theorem of Calculus.

Piecewise functions require careful handling, as the definition of the function changes at specific points. When finding intersection points or setting up integrals for areas involving piecewise functions, always consider the definitions over each specific interval.