Problem 9 - Thomas Calculus in SI Units

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{2}{3} x^{-1 / 3}\) b. \(\frac{1}{3} x^{-2 / 3}\) c. \(-\frac{1}{3} x^{-4 / 3}\)

Step-by-Step Solution

Answer

a. \(x^{2/3} + C\), b. \(x^{1/3} + C\), c. \(x^{-1/3} + C\)

Step 1: Understanding the Problem

We need to find the antiderivatives (indefinite integrals) for the given functions. Remember, an antiderivative is a function whose derivative gives back the original function.

Step 2: Applying the Power Rule for Integration

For functions of the form \(x^n\), the antiderivative is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration. This rule will be applied to each given function.

Step 3: Finding the Antiderivative for Part a

For \(\frac{2}{3} x^{-1 / 3}\), apply the power rule: \[\int \frac{2}{3} x^{-1/3} \, dx = \frac{2}{3} \cdot \frac{x^{2/3}}{2/3} + C = x^{2/3} + C\]

Step 4: Finding the Antiderivative for Part b

For \(\frac{1}{3} x^{-2 / 3}\), apply the power rule: \[\int \frac{1}{3} x^{-2/3} \, dx = \frac{1}{3} \cdot \frac{x^{1/3}}{1/3} + C = x^{1/3} + C\]

Step 5: Finding the Antiderivative for Part c

For \(-\frac{1}{3} x^{-4 / 3}\), apply the power rule: \[\int -\frac{1}{3} x^{-4/3} \, dx = -\frac{1}{3} \cdot \frac{x^{-1/3}}{-1/3} + C = x^{-1/3} + C\]

Step 6: Verification through Differentiation

Differentiate each found antiderivative to ensure it matches the original function. For example, differentiating \(x^{2/3} + C\) gives back \(\frac{2}{3} x^{-1/3}\), confirming correctness.

Key Concepts

Antiderivative

An antiderivative is essentially the reverse of a derivative. When you differentiate a function, you find its derivative, which represents its instantaneous rate of change. Conversely, the antiderivative is a function whose derivative equals the original function. In simpler terms, finding the antiderivative allows us to reconstruct the original function from its rate of change. This process is crucial in calculus, especially in integral calculus, because it helps in understanding how functions accumulate values over intervals. For example, sheathing the concept with our given exercise, the antiderivative of \(\frac{2}{3} x^{-1/3}\) is \(x^{2/3} + C\), where \(C\) is called the constant of integration. Remember, each function has many antiderivatives due to this constant \(C\). This reflects in the fact that differentiating any \(x^{2/3} + C\), regardless of what value \(C\) takes, will yield the same result: \(\frac{2}{3} x^{-1/3}\).
Understanding antiderivatives is key to solving problems related to area under curves and total change over intervals.

Power Rule for Integration

The power rule for integration is an important tool in finding antiderivatives. It's especially useful for functions of the form \(x^n\). This rule states that the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), as long as \(neq -1\). Here's a simple rundown:

• Increase the exponent \(n\) by 1.
• Divide by the new exponent \(n+1\).
• Add the constant of integration \(C\).

In the context of the given exercise, let's apply this rule to \(\frac{1}{3} x^{-2/3}\). By increasing \(-2/3\) by 1, we obtain \(1/3\). This gives us the antiderivative formula:\[\int \frac{1}{3} x^{-2/3} \, dx = \frac{1}{3} \cdot \frac{x^{1/3}}{1/3} + C = x^{1/3} + C\] This approach simplifies many integration problems and allows students to tackle them with confidence. The precision of this method hinges on correct exponent manipulation and constant inclusion.

Indefinite Integrals

Indefinite integrals are a core aspect of calculus because they represent all possible antiderivatives of a given function. Unlike definite integrals, which calculate a numerical value indicating the area under a curve between two points, indefinite integrals provide a general formula. This formula includes an arbitrary constant \(C\). The notation for an indefinite integral of a function \(f(x)\) is \(\int f(x) \, dx\).
To understand this concept, think of indefinite integrals as a reverse operation to differentiation. Our exercise involves finding the indefinite integral of specific functions. For instance, having found the antiderivative for \(-\frac{1}{3} x^{-4/3}\), the indefinite integral is:\[\int -\frac{1}{3} x^{-4/3} \, dx = x^{-1/3} + C\]Each function's antiderivative model includes \(C\) to account for the infinite possibilities of functions leading to the same derivative. Because of this nature, indefinite integrals themselves don't provide numerical answers, but they are instrumental when moving to more complex integration scenarios like definite integration.