Problem 40 - Calculus: Early Transcendental Functions

Question

Find the given area. The area below the \(x\) -axis and above \(y=x^{2}-4 x\)

Step-by-Step Solution

Answer

The area above the function \(y=x^{2}-4 x\) and below the \(x\)-axis between \(x = 0\) and \(x = 4\) is \(16/3\) units.

Step 1: Find the roots

First, let's find the roots of the function. This requires setting the function equal to zero and solving for \(x\). \(x^{2}-4 x=0\). Factoring out an \(x\) gives \(x*(x - 4) = 0\). Setting each factor equal to zero and solving for \(x\) gives us \(x = 0\) and \(x = 4\). These are the \(x\)-values where the graph intersects the \(x\)-axis.

Step 2: Set up the integral for the area

Now we set up the integral to find the area. Since we're looking for the area above the \(x\)-axis but below the curve, and the curve is continuous and non-negative in the interval [0,4], we look for the integral of the function from \(x = 0\) to \(x = 4\), which is \(\int_{0}^{4} (x^{2}-4 x) dx\).

Step 3: Solve the integral

Now, we solve the integral. We integrate term by term to get \([(1/3)x^3 - 2x^2]_{0}^{4}\). This integral evaluates to \((1/3) * 4^3 - 2 * 4^2\). Doing the arithmetic, we find that this equals \(16/3\) area units, which is the desired area above the graph \(y=x^{2}-4 x\) and below the \(x\)-axis within the interval [0,4].

Key Concepts

Definite Integral

When we talk about the definite integral in calculus, we are referring to a precise calculation of the accumulation of quantities. It can be visualized as the area under a curve between two points on the x-axis. In our exercise, we calculate the definite integral of the function from one point to another, which in this case are the roots of the function.

To set up a definite integral, you need to identify the limits of integration, which are the points where the function crosses the x-axis or other boundary points relevant to the problem. The area under the curve and above the x-axis for the function given, from the lower limit, which is 0, to the upper limit, which is 4, is represented by \[ \int_{0}^{4} (x^{2}-4 x) dx \. \] Solving this definite integral gives us the exact area contained within these bounds.

Area Under Curve

The 'area under the curve' is a concept that involves finding the size of the region under a graph of a function and above the x-axis. In many real-world scenarios, this would interpret as the total amount of a quantity changing over time. In integral calculus, the area under a curve is calculated using the definite integral, provided the function is continuous over the interval of interest.

In this instance, where our function dips below the x-axis, we consider the magnitude of the area without regard to whether the function lies above or below the x-axis. In other words, even though the function dips below the x-axis, the 'area under the curve' refers to the absolute value of this region. For quadratic functions, which are parabolas, the area can be between the curve and either the x-axis or the y-axis, depending on the orientation of the parabola.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, with the general form \( y = ax^2 + bx + c \). They have distinctive parabolic shapes that open upwards or downwards, depending on the sign of \( a \). Each parabola has a vertex, which is either the highest or lowest point on the graph, and can have either zero, one, or two x-intercepts—the points where it crosses or touches the x-axis.

In our exercise, we deal with the quadratic function \( y = x^2 - 4x \), which intercepts the x-axis at two points. These interception points actually represent the roots of the equation, found by setting \( y = 0 \), which are crucial in determining the interval for our definite integral. The shape and location of a quadratic curve are important when calculating the area under the curve, as they set the boundaries for the definitive integral calculation.