Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=x^{2}-4 x-5, y=0\), between \(x=-1\) and \(x=4\)

Short answer

The area is \(\frac{82}{3}\) or approximately 27.33 square units.

Step-by-step solution

Understand the Equations

We have two equations to consider: a quadratic function \(y = x^2 - 4x - 5\) and the line \(y = 0\), which is the x-axis. The region we are interested in is between \(x = -1\) and \(x = 4\). We are tasked with finding the area bounded by these graphs and the x-axis.

Determine Points of Intersection

To find the intersection points of \(y = x^2 - 4x - 5\) with the x-axis, set the quadratic function equal to zero: x^2 - 4x - 5 = 0. Solve this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1\), \(b=-4\), \(c=-5\). Solve it to find the roots, \(x_1 = -1\) and \(x_2 = 5\). The relevant region is from \(x = -1\) to \(x = 4\).

Sketch the Graph

Plot the graph of the quadratic function \(y = x^2 - 4x - 5\) over the interval \([-1, 4]\). Also, draw the x-axis (\(y=0\)). Notice that the curve intersects the x-axis at \(x = -1\) and \(x = 5\). We are interested in the area where the curve is below the x-axis between \(x = -1\) and \(x = 4\).

Set Up the Integral

The area under the curve from \(x = -1\) to \(x = 4\) can be represented as the integral of the function from \(-1\) to \(4\) since the function is below the x-axis (meaning the function value is negative, hence the need to integrate its negative or simply take the absolute value). Set up the integral: \[ A = \int_{-1}^{4} -(x^2 - 4x - 5) \, dx = \int_{-1}^{4} (-x^2 + 4x + 5) \, dx. \]

Calculate the Integral

Integrate the function: First, find the antiderivative: \[-\frac{x^3}{3} + 2x^2 + 5x\]. Evaluate this antiderivative from \(-1\) to \(4\): \[\left[-\frac{4^3}{3} + 2(4)^2 + 5(4)\right] - \left[-\frac{(-1)^3}{3} + 2(-1)^2 + 5(-1)\right].\]Calculate it step by step: \(-\frac{64}{3} + 32 + 20 = \frac{68}{3}\) for \(x=4\) and \(\frac{1}{3} + 2 - 5 = -\frac{14}{3}\) for \(x=-1\).Hence, the area is:\(\frac{68}{3} - \left(-\frac{14}{3}\right) = \frac{82}{3}.\)

Estimate the Area

For estimation, we can approximate some geometric shapes under the curve to cross-check the computed area. The curve is quadratic and symmetric; thus, it usually exhibits a parabolic shape. Use rough geometrical approximation, like combining triangles and rectangles, to ensure the calculated area of \(\frac{82}{3}\) which is approximately \(27.33\) square units closely matches the shape under the curve between \(-1\) and \(4\).

Key concepts

Definite Integral

The definite integral is a fundamental tool in calculus used to calculate the area under a curve over a certain interval. In the context of our exercise, it helps us find the area between the curve given by the quadratic function and the x-axis. This area is particularly interesting between the bounds from \(-1\) to \(4\).
When we perform definite integration, we evaluate the integral of a function over a specified interval. The result gives us the total net area between the curve and the x-axis. In our case, since the curve is below the x-axis in the interval, the integral needs to be taken as a negative, or we consider the absolute value of the definite integral, as only positive area is meaningful in this context.
Mathematically, for our specific problem, the definite integral is expressed as \[A = \int_{-1}^{4} -(x^2 - 4x - 5) \, dx \]This equation represents the sum of infinitely small areas under the curve between \(x = -1\) and \(x = 4\). Once the integral is calculated, the value gives us the total area under the curve. This is achieved by calculating the antiderivative of the function and then substituting the bounds into this antiderivative.

Quadratic Functions

Quadratic functions are a type of polynomial function with a degree of two, meaning they include an \(x^2\) term as the highest power of the variable. The general form of a quadratic function is given by \[f(x) = ax^2 + bx + c\]where \(a\), \(b\), and \(c\) are constants. In our exercise, the function is \[y = x^2 - 4x - 5\]
Quadratics graph as a parabola. The parabola opens upwards if the leading coefficient \(a\) is positive and downwards if \(a\) is negative. Here, \(a = 1\), so the parabola opens upwards.
This function has the fascinating property of being symmetric, meaning if you fold the parabola along the vertical line that passes through its vertex, both parts match perfectly. With quadratic functions, determining the team's vertex can offer insight into the graph's maximum or minimum value.
To sketch the graph, understanding the vertex and axis of symmetry is helpful. However, for this particular problem, we are only concerned about where the parabola intersects the x-axis, which we call the roots or the solutions. These intersections help form the region whose area we are analyzing.

Intersection Points

Intersection points are where two graphs meet or cross each other. In the context of our exercise, we look for where the quadratic curve intersects the line defined by \(y=0\), which is the x-axis. These points are crucial in finding the area under the curve because they define the bounds of the region for which we calculate the area.
To identify these points, set the quadratic equation equal to zero and solve for \(x\): \[x^2 - 4x - 5 = 0\]
We solve using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Given \(a = 1\), \(b = -4\), and \(c = -5\), we find the solutions to be \(x_1 = -1\) and \(x_2 = 5\).
These are our exact intersection points with the x-axis. However, for our problem, since we're considering only the interval from \(-1\) to \(4\), the relevant intersection occurs at \(x = -1\). The other intersection at \(x = 5\) is outside the range we're looking at, so it's not considered for the region bounded by the given limits. This interval sets the stage for our area analysis between the curve and the x-axis.