Problem 83 - Algebra and Trigonometry

Question

Find the area bounded by the graphs of \(y=\frac{3}{5} x+\frac{12}{5}, y=-x+4,\) and \(y=-\sqrt{16-x^{2}}\)

Step-by-Step Solution

Answer

The area is determined through defined integrals and subtraction yielding fixed bounds within intersections.

Step 1 - Identify Points of Intersection

First, find the points where the curves intersect each other. Solve the equations \(\frac{3}{5} x + \frac{12}{5} = -x + 4\) and \(\frac{3}{5} x + \frac{12}{5} = -\sqrt{16 - x^2}\) and \(-x + 4 = -\sqrt{16 - x^2}\).

Step 2 - Solve for \(\frac{3}{5} x + \frac{12}{5} = -x + 4\)

This simplifies to \(\frac{8}{5} x + \frac{12}{5} = 4\). By solving for \(x\), obtain \(x = 1\). Substitute \(x = 1\) back into the equation \(y = -x + 4\) to find \(y = 3\). Thus, the intersection point is \((1, 3)\).

Step 3 - Solve for \(\frac{3}{5} x + \frac{12}{5} = -\sqrt{16 - x^2}\)

This equation requires solving simultaneously for both sides; however, observe that \(y = -\sqrt{16 - x^2}\) defines a semicircle which intersects only at specific points. Calculations simplify, leading to \(x = -4\) or \(4\) where valid intersections occur within limits.

Step 4 - Solve for \(-x + 4 = -\sqrt{16 - x^2}\)

This has solutions where \(x\) gives valid points by simplifying to intersection contours. The result is \((0, 4)\).

Step 5 - Define Integration Limits

Determine integration limits for area calculation where the functions of \(y = \frac{3}{5} x + \frac{12}{5}, y = -x + 4, y = -\sqrt{16 - x^2}\) overlap. These are bounded within x-values satisfying intersections.

Step 6 - Set Up Integral Expression

Integrate the differences of the upper and lower functions bound within the limits: \(\text{Area} = \ \int_{a}^{b} (\frac{3}{5}x + \frac{12}{5} - (-x + 4)) dx - \ \int_{c}^{d} (-x + 4 + \sqrt{16-x^2}) dx\).

Step 7 - Evaluate the Integral

Compute the definite integrals using calculated limits:\[\int_{-4}^{1} (\frac{8}{5}x + \frac{8}{5}) dx - \int_{1}^{0} (-x + 4 + \sqrt{16-x^2}) dx\]. Simplifying each integral component for cumulative area results.

Key Concepts

Area Between Curves

Finding the area between curves involves integrating the difference between the upper and lower functions across a given interval. First, determine which function is on top and which is on the bottom. This helps to set up the integral properly.
For example, if we have curves given by functions \(f(x)\) and \(g(x)\) over an interval \([a, b]\), where \(f(x) \ge g(x)\), then the area between the curves is found using:
\[\text{Area} = \int_{a}^{b} (f(x) - g(x)) dx\]
This concept is crucial for accurately calculating the area and ensures that values are positive. Mistakes in identifying top and bottom functions can lead to errors in the calculated area.

Integration

Integration is the process of finding the area under a curve. When dealing with areas between curves, we use definite integrals. In a definite integral, we are interested in the area under the curve between two specific points.
Unlike indefinite integrals, which include a constant of integration, definite integrals represent exact areas.
To integrate a function \(f(x)\) from \(a\) to \(b\), we compute \[\text{Integral} = \int_{a}^{b} f(x) dx\]
This gives us the total area under the function \(f(x)\) from \(x=a\) to \(x=b\). For areas between curves, consider the difference between the upper and lower functions.

Points of Intersection

Points of intersection are where two or more curves meet. Finding these points is essential for setting integration limits.
To find points of intersection, solve the equations of the curves simultaneously. For instance, given curves \(y = f(x)\) and \(y = g(x)\), set \(f(x) = g(x)\) and solve for \(x\). Substituting these \(x\) values back into one of the original equations provides the corresponding \(y\) values.
In the example problem, curves intersect where:

• \(\frac{3}{5}x + \frac{12}{5} = -x + 4\)
• \(\frac{3}{5}x + \frac{12}{5} = -\sqrt{16 - x^2}\)
• \(-x + 4 = -\sqrt{16 - x^2}\)

From these, we determine the limits for our integrals.

Definite Integral

The definite integral \(\int_{a}^{b} f(x) dx\) represents the exact area under the curve \(f(x)\) from \(x=a\) to \(x=b\).
It's calculated using the antiderivative \(F(x)\) of \(f(x)\), and applying the limits \(a\) and \(b\). This is given by the Fundamental Theorem of Calculus: \[{F(b) - F(a)} = \int_{a}^{b} f(x) dx\]
For the given problem:
Integrate from \(x = -4\) to \(x = 1\) for the first area and from \(x = 1\) to \(x = 0\) for the second area. The setup for evaluating becomes: \[\text{Area} = \int_{-4}^{1} \left( \frac{8}{5}x + \frac{8}{5} \right) dx - \int_{1}^{0} \left( -x + 4 + \sqrt{16 - x^2} \right) dx\]
By solving these, we get the exact area bounded by the curves. This process illustrates the power of definite integrals for determining bounded areas between intersecting curves.