Question

Use the most efficient strategy for computing the area of the following regions. The region bounded by \(y=x^{2}-4,4 y-5 x-5=0,\) and \(y=0,\) for \(y \geq 0\)

Short answer

Question: Find the area of the region bounded by the equations \(y = x^2 - 4\), \(4y - 5x - 5 = 0\), and \(y = 0\). Answer: The area of the region is approximately \(18.87\).

Step-by-step solution

Find the points of intersection

First, let's solve for the intersections between each pair of curves. The given equations are: 1. \(y = x^2 - 4\) 2. \(4y - 5x - 5 = 0\) 3. \(y = 0\) Intersections between (1) and (2): Set \(y = x^2 - 4\) in equation (2) \(4(x^2 - 4) - 5x - 5 = 0 \Rightarrow 4x^2 - 5x - 21 = 0\) This is a quadratic equation, whose solutions are \(x_1 = 3\) and \(x_2 = -\frac{7}{4}\). Corresponding y values are \(y_1 = 5\) and \(y_2 = -\frac{15}{16}\). Intersections between (1) and (3): Set \(y = 0\) in equation (1) \(0 = x^2 - 4 \Rightarrow x = \pm 2\). Since we are only considering positive y values \((y \geq 0)\), we pick \(x = 2\). Then the intersection point is \((2,0)\). Intersections between (2) and (3): Set \(y = 0\) in equation (2) \(4(0) - 5x - 5 = 0 \Rightarrow x = -1\). Since we are only considering positive y values \((y \geq 0)\), we pick the intersection at \(y = 0\), which is point \((2,0)\) that we found earlier. From the intersection points, we find that our area of interest is situated between the curves of equations (1) and (2), between the points \((-\frac{7}{4}, -\frac{15}{16})\) and \((3, 5)\), and above the x-axis.

Set up the integral

Now, we will find the area by integrating along the x-axis between the intersection points. First, we need to solve equation (2) for y: \(4y - 5x - 5 = 0 \Rightarrow y = \frac{5}{4}x + \frac{5}{4}\) Next, we will subtract the equation of curve (1) from curve (2). This gives us the height of the region at any point x, and the integral represents the accumulation of these heights: \(\int_{-\frac{7}{4}}^{3} \left(\frac{5}{4}x + \frac{5}{4} - (x^2 - 4)\right) dx\)

Evaluate the integral

Now, we will evaluate the integral to find the area: \(\int_{-\frac{7}{4}}^{3} \left(\frac{5}{4}x + \frac{5}{4} - (x^2 - 4)\right) dx = \int_{-\frac{7}{4}}^{3} \left(-x^2 + \frac{5}{4}x + 9\right) dx\) As the integral is a polynomial, we can find its antiderivative easily: \(F(x) = \left[-\frac{1}{3} x^3 + \frac{5}{8} x^2 + 9x\right]_{-\frac{7}{4}}^{3}\) Now, we plug in the limits of integration and calculate the difference: \(F(3) - F\left(-\frac{7}{4}\right) = \left(-\frac{1}{3} (3)^3 + \frac{5}{8} (3)^2 + 9(3)\right) - \left(-\frac{1}{3} \left(-\frac{7}{4}\right)^3 + \frac{5}{8} \left(-\frac{7}{4}\right)^2 + 9\left(-\frac{7}{4}\right)\right)\) After simplifying the expression, we get the area of the region: \(Area \approx 18.87\)

Key concepts

Definite Integral

In calculus, a **definite integral** is a crucial concept used to determine the accumulation of quantities, such as areas under curves. It is represented as the integral of a function over a specified interval. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted by \[ \int_{a}^{b} f(x) \, dx. \]Here, \( a \) and \( b \) are the limits of integration, defining the range over which we find the accumulation.
The process of finding a definite integral involves determining the antiderivative of the function, which is then evaluated at the upper and lower limits. The difference gives the area (or other accumulations) between the curve and the x-axis along the interval.
Definite integrals are central to solving problems related to the area between curves, which involves computing the difference in heights of two functions over a specified range. By evaluating this integral, you determine the "net area" within that interval, accommodating both positive and negative areas generated by the function spectrums.

Quadratic Equation

A **quadratic equation** is a fundamental algebraic expression in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The highest exponent of the variable \( x \) is 2, making it a second-degree polynomial equation.
Quadratic equations can be solved using various methods, such as:

• Factoring
• Completing the square
• Using the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)

Solving a quadratic equation gives the roots or solutions (\( x \)-values) where the quadratic function meets the x-axis.
These solutions, or intersections, help in determining points of interest in many calculus applications, such as where a curve intersects other curves or axes. In our context, finding the intersections involved solving the quadratic equation derived from setting the equations of the bounding curves equal to each other.

Points of Intersection

In geometry and calculus, **points of intersection** are the coordinates (\( x, y \)) where two or more graphs meet. These points are crucial for identifying regions bounded by multiple functions.
To find points of intersection, set the equations of two functions equal and solve for the variable. The solutions indicate the x-values where the curves or lines intersect. For each x-value, substitute back into one of the original equations to find the corresponding y-value.
Points of intersection are important when calculating areas enclosed by curves, as they define the limits of integration used in definite integrals. These points serve as vertices of the region of interest, helping to determine the segment over which we perform integration. Understanding these intersections ensures accurate analysis and calculations when dealing with various geometric shapes formed by multiple functions.