Problem 10 - Technical Mathematics with Calculus

Question

Find the area bounded by the curve \(y=x^{3},\) the \(x\) axis, and the lines \(x=-3\) and \(x=0\)

Step-by-Step Solution

Answer

\frac{81}{4} \text{ square units}

Step 1: Sketch the Graph

Firstly, plot the graph of the function \(y = x^3\) and identify the area under consideration. You need to sketch the curve between \(x=-3\) and \(x=0\) and determine the region that lies above the x-axis and below the curve within these limits.

Step 2: Identify Bounds for Integration

To find the area, set up the integral with the appropriate bounds. Since we're looking for the area between the curve and the x-axis from \(x=-3\) to \(x=0\), our limits of integration are -3 and 0.

Step 3: Set Up the Integral

Write the integral that represents the area under the curve \(y=x^3\). The integral will be \(\int_{-3}^{0} x^3 dx\). This will calculate the total area under the curve from \(x=-3\) to \(x=0\).

Step 4: Evaluate the Integral

Evaluate the integral by calculating the antiderivative of \(x^3\), which is \(\frac{x^4}{4}\). After finding the antiderivative, apply the upper and lower bounds of the integral to find the area. This yields \(\left.\frac{x^4}{4}\right|_{-3}^{0} = \frac{0^4}{4} - \frac{(-3)^4}{4}= -\frac{81}{4}\).

Step 5: Calculate the Absolute Value of the Result

Since area cannot be negative, take the absolute value of the result obtained from the integral. The absolute value of \(-\frac{81}{4}\) is \(\frac{81}{4}\). This gives us the final area bounded by the curve, the x-axis, and the vertical lines \(x=-3\) and \(x=0\).

Key Concepts

Definite Integral

In calculus, a definite integral is used to calculate the area under a curve between two points on the x-axis. To understand this, imagine drawing a curve on a graph and shading the area beneath it, from one specified point to another. This 'shaded region' can often represent physical concepts like displacement or the accumulated quantity of a substance.

In our exercise, the definite integral is represented as \(\int_{-3}^{0} x^3 dx\). The integral sign \(\int\) indicates the operation of integration, while the numbers -3 and 0 are the lower and upper bounds, respectively. These numbers tell you where to start and stop measuring the area. By evaluating this integral, we can find the exact space enclosed by the curve \(y=x^3\) and the x-axis, bounded by the vertical lines at \(x=-3\) and \(x=0\).

Integration Bounds

The 'integration bounds' or 'limits of integration' are crucial in the process of finding a definite integral. They set the endpoints of the interval over which the function is being integrated. The lower bound of the integral indicates the starting point, and the upper bound shows the end point of the interval on the x-axis.

In our case, the bounds are given by \(x=-3\) and \(x=0\). These numbers dictate the section of the curve where we are focusing our attention. It is worth noting that if we reverse these bounds, the result from the integral would change sign, emphasizing the importance of the correct order. Bounds must always be evaluated properly to ensure that the integral accurately reflects the area being measured.

Antiderivative

The antiderivative, in the simplest terms, is the reverse process of differentiation. While differentiation gives you the rate at which something changes, the antiderivative helps you find the original quantity if you know the rate of change.

In the context of integrals, the antiderivative is the function you get after performing integration, essentially undoing a derivative. For the example given, \(x^3\) is integrated to give us its antiderivative \(\frac{x^4}{4}\), which is then used to calculate the area. Understanding how to find and apply antiderivatives is a fundamental part of solving definite integrals because it allows us to evaluate the integral at our specified bounds.

Absolute Value in Integration

The concept of absolute value in integration comes into play when the function or its graph lies below the x-axis over the interval we are interested in. Since the area cannot be negative, we use the absolute value to ensure we get a positive measure of space.

In the provided exercise, after evaluating the antiderivative with the integral bounds, we got a negative result \( -\frac{81}{4}\), which refers to the area lying below the x-axis. However, in the context of finding 'actual' area, we interpret this as a positive space by taking the absolute value, giving us \( \frac{81}{4}\). Always remember, when computing areas using definite integrals, to apply the absolute value to negative results to make sense of them in a physical context.