Problem 15 - Precalculus: Mathematical for Calculus

Question

Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=x^{3}+2, \quad 0 \leq x \leq 5$$

Step-by-Step Solution

Answer

The area under the graph is 166.25 square units.

Step 1: Understand the Problem

To find the area under the curve of a function over an interval, we use definite integration. The function given is \(f(x) = x^3 + 2\) and the interval is \([0, 5]\). This involves evaluating \(\int_{0}^{5} (x^3 + 2)\, dx\).

Step 2: Set Up the Integral

Write the integral that represents the area under the curve. We have:\[\int_{0}^{5} (x^3 + 2)\, dx.\]

Step 3: Apply the Power Rule

To integrate \(x^3 + 2\), apply the power rule of integration: \(\int x^n\, dx = \frac{x^{n+1}}{n+1} + C\). Therefore:\[\int (x^3)\, dx = \frac{x^4}{4}.\] Also, for the constant \(2\), \(\int 2\, dx = 2x\). Thus:\[\int (x^3 + 2)\, dx = \frac{x^4}{4} + 2x.\]

Step 4: Evaluate the Definite Integral

Using the Fundamental Theorem of Calculus, evaluate the definite integral from 0 to 5:\[\left[\frac{x^4}{4} + 2x\right]_{0}^{5} = \left(\frac{5^4}{4} + 2(5)\right) - \left(\frac{0^4}{4} + 2(0)\right).\]

Step 5: Compute the Values

Calculate \(5^4 = 625\), so \(\frac{625}{4} = 156.25\). Also, calculate \(2 \times 5 = 10\). The integral evaluates to:\[156.25 + 10 = 166.25.\] The lower limit evaluates to 0, which doesn't affect the result. So, the area is 166.25.

Key Concepts

Power Rule of Integration

When integrating functions, especially polynomials, one of the most useful tools is the Power Rule of Integration. It's a straightforward method that can simplify the process significantly. The rule states that the integral of a function in the form of \(x^n\) is \( \int x^n \; dx = \frac{x^{n+1}}{n+1} + C \). Here, \(C\) is the constant of integration, which is typically omitted in definite integrals because we evaluate at specific points.

For the exercise given, integrating \(x^3 + 2\) uses this rule. The \(x^3\) term becomes \(\frac{x^4}{4}\) when integrated, because we increase the power by 1 (from 3 to 4) and divide by this new power (4). This calculation is a vital step in solving the problem because it provides the structure for evaluating the definite integral over the interval \([0, 5]\).

The power rule makes it easy to break complex polynomial functions into simpler steps. This approach helps us find areas, among other calculations, by making integration more manageable.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is one of the cornerstones that link the concepts of differentiation and integration. It tells us that integration can be seen as the inverse operation to differentiation. Specifically, it helps us evaluate definite integrals with ease.

The theorem has two main parts.

• The first part connects a function's antiderivative with its definite integral.
• The second part states that if \(F\) is an antiderivative of \(f\), then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

In the exercise, we apply this theorem to evaluate the integral \(\int_{0}^{5} (x^3 + 2) . dx\).

After finding the antiderivative \(\frac{x^4}{4} + 2x\), we use the theorem to calculate \(\left[\frac{5^4}{4} + 2 \times 5\right] - \left[\frac{0^4}{4} + 2 \times 0\right]\). The definite limits turn the process into a simple subtraction of evaluations, making it efficient to compute the area under the curve.

Area under a curve

In calculus, the concept of 'area under a curve' is central to understanding how integration works in practical terms. By finding the area, we essentially sum up an infinite number of infinitesimally small slices under the function over a given interval.

For the function \(f(x) = x^3 + 2\) considered in this exercise, we're interested in the interval between \(x = 0\) and \(x = 5\). Calculating this area involves performing a definite integral, which provides a precise measure of the space beneath the curve of the function and above the x-axis, between these two points.

Definite integration calculates total accumulated values, such as areas, by considering the net value. In practical terms, this means when evaluating \(\int_{0}^{5} (x^3 + 2)\, dx\), we find the total area by evaluating the bounds. This method breaks the continuous function into understandable pieces, which when added up, give the precise measure of the area, such as the result of 166.25 in this case.

Understanding the area under the curve aids in many real-world applications, from calculating distances to evaluating various physical phenomena.