Problem 41 - Applied Calculus

Question

a. Verify that \(\int x^{2} d x=\frac{1}{3} x^{3}+C\). b. Graph the five functions \(\frac{1}{3} x^{3}-2, \frac{1}{3} x^{3}-1\), \(\frac{1}{3} x^{3}, \frac{1}{3} x^{3}+1,\) and \(\frac{1}{3} x^{3}+2\) (the solutions for five different values of \(C\) ) on the window [-3,3] by \([-5,5] .\) Use TRACE to see how the constant shifts the curve vertically. c. Find the slopes (using NDERIV or \(d y / d x\) ) of several of the curves at a particular \(x\) -value and check that in each case the slope is the square of the \(x\) -value. This verifies that the derivative of each curve is \(x^{2}\), and so each is an integral of \(x^{2}\).

Step-by-Step Solution

Answer

The integral is verified; graphs illustrate vertical shifts; slopes are consistent with \(x^2\).

Step 1: Verify the Integral

To verify that \( \int x^{2} \, dx = \frac{1}{3} x^{3} + C \), take the derivative of \( \frac{1}{3} x^{3} + C \) with respect to \( x \). The derivative \( \frac{d}{dx} \left(\frac{1}{3} x^{3} + C \right) \) is \( x^{2} \), as the constant \( C \) differentiates to zero and the power rule \( \frac{d}{dx} x^n = n x^{n-1} \) is applied, resulting in \( 3 \times \frac{1}{3} x^{3-1} = x^{2} \). This confirms the integral correct.

Step 2: Sketch Graphs for Different Constants

Graph the functions \( \frac{1}{3} x^{3} - 2, \frac{1}{3} x^{3} - 1, \frac{1}{3} x^{3}, \frac{1}{3} x^{3} + 1, \frac{1}{3} x^{3} + 2 \) in the window \([-3, 3]\) by \([-5, 5]\). Use a graphing tool or software. Each curve will have the same shape but shifted vertically based on its constant \(C\). This illustrates the family of antiderivatives for \( x^2 \).

Step 3: Analyze Vertical Shifts Using TRACE

Use the TRACE feature in the graphing tool to observe that changing the constant \(C\) shifts the graph up or down. The function \( \frac{1}{3} x^{3} + C \) gets translated vertically without affecting its slope.

Step 4: Calculate Slopes of Functions

Use a calculator or software to calculate the derivative (slope) of the function \( \frac{1}{3} x^{3} + C \) at different \( x \)-values (e.g., \( x = -2, 0, 2 \)). The derivative function is \( x^{2} \), so at any point \( x \), the slope should be \( x^{2} \). For example, at \( x = 2 \), the slope is \( 4 \), and at \( x = -1 \), the slope is \( 1 \).

Step 5: Verify Consistency of Integral

Check that for each value of \( x \), the calculated slopes (from Step 4) match the square of \( x \). This confirms that the derivative of each function is \( x^{2} \), thus verifying that each function is indeed an integral of \( x^{2} \).

Key Concepts

Antiderivatives

An antiderivative is a function whose derivative is the original function. It is the reverse process of differentiation. In calculus, finding an antiderivative means solving an indefinite integral. When you solve \[ \int f(x) \, dx \]you find a function such that when differentiated, it results in \( f(x) \).A handy way to think about it is to imagine the process of undoing a derivative. For example, the function \( \frac{1}{3}x^3 + C \) is an antiderivative of \( x^2 \). When you differentiate \( \frac{1}{3}x^3 + C \), you get \( x^2 \), confirming the process. This often includes a constant \( C \) because differentiation of any constant is zero, making an entire family of antiderivatives possible for a given function.

Power Rule

The Power Rule is a fundamental tool used in calculus for both differentiation and integration. It provides a quick and easy way to find derivatives and integrals of functions of the form \( x^n \).For differentiation, the Power Rule states:\[\frac{d}{dx} (x^n) = n x^{n-1}\]This means that when you take the derivative of \( x^n \), you bring the exponent down as a multiplier and subtract one from the original exponent.For integration, the Power Rule is slightly modified:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Here, to find the integral (antiderivative), you add one to the original exponent and divide by the new exponent, then add an arbitrary constant \( C \). This adjustment provides the general form for indefinite integrals.

Graphing Functions

Graphing functions provides a visual representation of mathematical relationships and can help in understanding changes and behaviors in functions. When graphing the family of functions \( \frac{1}{3}x^3 + C \) based on different values of \( C \), you can observe the impact of this constant. Each graph of \( \frac{1}{3}x^3 + C \) has the same cubic shape but is vertically shifted:

• \( \frac{1}{3}x^3 - 2 \)
• \( \frac{1}{3}x^3 - 1 \)
• \( \frac{1}{3}x^3 \)
• \( \frac{1}{3}x^3 + 1 \)
• \( \frac{1}{3}x^3 + 2 \)

By graphing these on a coordinate plane, you can see the effect of the constant \( C \). It shifts the entire curve up or down, but it doesn't change the basic form or slope of the curve.

Derivatives

Derivatives represent the rate at which a function changes. They are used to find slopes of curves at given points, making them essential in calculus. Specifically, the derivative of a function shows how it increases or decreases at any given point \( x \).In our exercise, the derivative of \( \frac{1}{3}x^3 + C \) is calculated as \( x^2 \), aligning with the original integrand. Calculating slopes using the derivative function confirms the relationship between differentiation and integration. Use these steps to compute the derivative:

• Apply the Power Rule to each term.
• Check computations at various \( x \) values to ensure accuracy.
• At any point \( x \), the slope should equal \( x^2 \).

Verifying these slopes confirms the function properly reverses the integration process. For instance, when \( x = 2 \), the slope \( x^2 \) is \( 4 \), as expected, ensuring the correct function behavior.