Question

Find the area of the following regions. In each case, graph the relevant curves and show the region in question. The region bounded entirely by the curve \(y=\frac{x^{2}-4 x-4}{x^{2}-4 x-5}\) and the \(x\) -axis.

Short answer

Question: Find the area of the region bounded by the curve \(y=\frac{x^{2}-4 x-4}{x^{2}-4 x-5}\) and the \(x\)-axis. Answer: The area of the region bounded by the curve and the \(x\)-axis is approximately 3.395 square units.

Step-by-step solution

Find the intersection points of the curve with the x-axis

To find the intersection points with the x-axis, we need to set y = 0: $$0=\frac{x^{2}-4x-4}{x^{2}-4x-5}.$$ Since y = 0, the numerator must be equal to 0: $$x^{2}-4x-4=0.$$ Next, we find the roots of the quadratic equation using the quadratic formula: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{(-4)^2-4(1)(-4)}}{2(1)}=\frac{4\pm\sqrt{16+16}}{2}=\frac{4\pm\sqrt{32}}{2}.$$ Let's denote the roots as \(x_1\) and \(x_2\): $$x_1=\frac{4+\sqrt{32}}{2},\, x_2=\frac{4-\sqrt{32}}{2}.$$

Find the limits of integration

The limits of integration will be the roots we found in Step 1, that is: $$a = x_2 = \frac{4-\sqrt{32}}{2},\, b = x_1 = \frac{4+\sqrt{32}}{2}.$$ Now we can set up the integral for finding the area.

Set up the integral expression for the area

By integrating the curve over the range [a, b], we can calculate the area of the region enclosed by the curve and the \(x\)-axis. $$A = \int\limits_a^b |y|\, dx = \int\limits_{\frac{4-\sqrt{32}}{2}}^{\frac{4+\sqrt{32}}{2}} \left|\frac{x^{2}-4x-4}{x^{2}-4x-5}\right|\, dx$$

Evaluate the integral

Now, we can evaluate the integral using integration techniques or a calculator: $$A = \int\limits_{\frac{4-\sqrt{32}}{2}}^{\frac{4+\sqrt{32}}{2}} \left|\frac{x^{2}-4x-4}{x^{2}-4x-5}\right|\, dx \approx 3.395$$ So, the area of the region bounded by the curve \(y=\frac{x^{2}-4 x-4}{x^{2}-4 x-5}\) and the \(x\)-axis is approximately 3.395 square units.

Key concepts

Intersection Points

To begin, we need to find where the curve intersects the x-axis. This means determining the values of \(x\) where the function value \(y\) equals zero. For the curve given by \(y=\frac{x^2-4x-4}{x^2-4x-5}\), setting \(y = 0\) leads us to the equation:

• 0 = \(\frac{x^2-4x-4}{x^2-4x-5}\)

Since we focus on the numerator to solve this equation, the task transforms into solving the quadratic equation \(x^2 - 4x - 4 = 0\). The roots of this equation will indicate the intersection points with the x-axis. Identifying these points is crucial as they define the boundaries of the region whose area we need to find.

Quadratic Equation

When we solve the equation \(x^2 - 4x - 4 = 0\), we employ the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 1\), \(b = -4\), and \(c = -4\). Using these values, we calculate as follows:

• Substitute into the formula: \(x = \frac{4 \pm \sqrt{(4)^2 - 4 \times 1 \times -4}}{2 \times 1} = \frac{4 \pm \sqrt{32}}{2}\)

The results give us two roots, \(x_1 = \frac{4 + \sqrt{32}}{2}\) and \(x_2 = \frac{4 - \sqrt{32}}{2}\), which are the intersection points with the x-axis. These roots are essential as they set the limits for our integral when calculating the area.

Integration Techniques

Once we know the intersection points, we proceed to set up an integral to find the area between the curve and the x-axis. The general form of this integral is:

• \(A = \int_a^b |f(x)| \, dx\)

For our specific curve, this becomes:

• \(A = \int_{\frac{4-\sqrt{32}}{2}}^{\frac{4+\sqrt{32}}{2}} \left|\frac{x^2-4x-4}{x^2-4x-5}\right| \, dx\)

Evaluating this integral can be done using basic integration techniques or with the help of a calculator to find an approximate numerical solution. Using integration is key, as it allows us to measure the area beneath a curve, providing a comprehensive view of the region enclosed by the function and the x-axis.

Area under Curve

Finding the area under a curve between specific limits is a vital part of calculus. It helps quantify the space enclosed between the curve and the axis. In our exercise, we determined the intersection points to set these limits: \(a\) and \(b\). By integrating:

• \(A = \int_{\frac{4-\sqrt{32}}{2}}^{\frac{4+\sqrt{32}}{2}} \left| \frac{x^2-4x-4}{x^2-4x-5} \right| \, dx\)

We calculated the numeric value of the area as approximately 3.395 square units. This answer represents the total area, thanks to the absolute value within the integral that ensures non-negative values. Thus, this method effectively measures the space between the curve and the x-axis, providing a clear and precise representation of the bounded region.