Problem 12 - Applied Calculus

Question

(a) Use a graph of the integrand to make a rough estimate of the integral. Explain your reasoning. (b) Use a computer or calculator to find the value of the definite integral. $$ \int_{0}^{1} x^{3} d x $$

Step-by-Step Solution

Answer

The rough estimate is about 0.25. The definite integral is exactly 0.25.

Step 1: Understand the Integrand

The integrand is the function we are integrating, which is \(x^3\). This function is a simple polynomial with a degree of 3, which means it has a curve that starts from zero on the x-axis and increases as x increases.

Step 2: Sketch the Graph

Plot the graph of \(y = x^3\) from \(x = 0\) to \(x = 1\). The graph should start at the origin (0,0) and curve upwards to (1,1). Since it is a cubic function, it increases at an increasing rate as \(x\) approaches 1.

Step 3: Make a Rough Estimate of the Area

The definite integral \(\int_{0}^{1} x^3 \, dx\) is the area under the curve from \(x = 0\) to \(x = 1\). Observing the plot, notice that the area is less than a triangle with a base and height of 1. A rough estimate can be around 0.25 to 0.5, but closer to 0.25.

Step 4: Setup the Definite Integral

Set up the definite integral \(\int_{0}^{1} x^3 \, dx\). The integral represents the exact area under the curve from 0 to 1.

Step 5: Calculate the Definite Integral

To find the exact value, use the antiderivative: \(F(x) = \frac{x^4}{4}\). Apply the Fundamental Theorem of Calculus: \[\text{Evaluate } \left[\frac{x^4}{4}\right]_{0}^{1} \] Calculate: \[= \frac{(1)^4}{4} - \frac{(0)^4}{4} = \frac{1}{4} - 0 = \frac{1}{4}\]

Step 6: Verify with a Calculator

Use a calculator or computer to compute \(\int_{0}^{1} x^3 \, dx\) to verify our manual calculation. The computed value should be 0.25, confirming our calculation.

Key Concepts

Polynomial Integration

Polynomial integration involves finding the antiderivative of a polynomial function. In the exercise, the integrand is a cubic polynomial, specifically, \(x^3\).
Integrating polynomials follows a straightforward process, where for any term \(x^n\), the antiderivative is \(\frac{x^{n+1}}{n+1}\).
For example, to integrate \(x^3\), the antiderivative is \(\frac{x^4}{4}\). This means to find the definite integral, you use the antiderivative to evaluate the area under the curve across the specified interval, from \(0\) to \(1\) in this case.

• First, determine the antiderivative: \(F(x) = \frac{x^4}{4}\).
• Then, evaluate the function at the boundaries of the integral.
• Finally, find the difference: \([F(1) - F(0)]\).

Polynomial integration is a key skill as it provides a method to calculate exact areas under curves formulated as polynomial functions.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) links the concept of differentiation and integration, establishing a foundational principle in calculus.
In practice, it simplifies the computation of a definite integral if the antiderivative is known. The theorem states: If \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
This principle allows you to evaluate the definite integral in the exercise \(\int_{0}^{1} x^3 \, dx\).
To apply it:

• Find the antiderivative, \(F(x) = \frac{x^4}{4}\).
• Evaluate at the bounds: \(F(1) = \frac{1}{4}\) and \(F(0) = 0\).
• Subtract these values: \( \frac{1}{4} - 0 = \frac{1}{4} \).

Using the FTC, you effectively transition from finding indefinite integrals to calculating exact values for definite integrals.

Graphical Estimation of Integrals

Graphical estimation involves visualizing the function to estimate the integral's value. This process gives a rough idea of the area under the curve.
In the exercise, you are asked to graph \(y = x^3\) between \(x = 0\) and \(x = 1\), and estimate the area under this curve.

• Sketch how the function increases from the origin to (1,1).
• Recognize that the shape of the \(x^3\) curve is not linear, but curves upward with an increasing slope.
• Estimate the area under the curve, comparing it to more familiar shapes, like a triangle with base and height of 1.

This visual method is useful as it provides intuitive understanding, confirming further computations.
While not as precise as algebraic methods, graphical estimation is a handy tool to approximate integrals quickly.