Problem 4 - Essential Calculus Early Transcendentals

Question

Find the most general antiderivative of the function. (Check your answer by differentiation.) $$f(x)=\sqrt[3]{x^{2}}+x \sqrt{x}$$

Step-by-Step Solution

Answer

The antiderivative is \( F(x) = \frac{3}{5}x^{5/3} + \frac{2}{5}x^{5/2} + C \).

Step 1: Rewrite the Function with Exponents

We start by rewriting the given function using exponents. The function is \( f(x) = \sqrt[3]{x^2} + x \sqrt{x} \). In terms of exponents, this becomes \( f(x) = x^{2/3} + x^{3/2} \).

Step 2: Find the Antiderivative of Each Term

To find the most general antiderivative, integrate each term individually:1. For the term \( x^{2/3} \), the antiderivative is \( \frac{3}{5}x^{5/3} \) (add 1 to the exponent and divide by the new exponent).2. For the term \( x^{3/2} \), the antiderivative is \( \frac{2}{5}x^{5/2} \).The general antiderivative is \[ F(x) = \frac{3}{5}x^{5/3} + \frac{2}{5}x^{5/2} + C, \]where \( C \) is an arbitrary constant.

Step 3: Differentiate the Antiderivative to Verify

Differentiate the antiderivative \( F(x) = \frac{3}{5}x^{5/3} + \frac{2}{5}x^{5/2} + C \).Using the power rule:1. The derivative of \( \frac{3}{5}x^{5/3} \) is \( x^{2/3} \).2. The derivative of \( \frac{2}{5}x^{5/2} \) is \( x^{3/2} \).Thus, the derivative \( F'(x) = x^{2/3} + x^{3/2} \) which matches the original function \( f(x) \).

Key Concepts

Power Rule for Integration

The power rule is a fundamental technique used in calculus to find antiderivatives, which are the reverse of derivatives. When integrating terms that follow the form \( x^n \), where \( n eq -1 \), the power rule involves the following steps:

• Increase the exponent by one, transforming it from \( n \) to \( n+1 \).
• Divide the term by the new exponent \( n+1 \).

In mathematical terms, the integral \( \int x^n \, dx \) leads to the antiderivative \( \frac{x^{n+1}}{n+1} + C \), where \( C \) represents an arbitrary constant.

For example, if we take the term \( x^{2/3} \), applying the power rule gives the antiderivative \( \frac{3}{5}x^{5/3} \). Similarly, for \( x^{3/2} \), its antiderivative is \( \frac{2}{5}x^{5/2} \). These simple calculations illustrate the elegance and efficiency of using the power rule in finding antiderivatives.

Understanding Exponentials

Exponential functions, although often associated with expressions like \( e^x \), also appear in forms involving variables raised to a fractional power, much like our example equation.

In the problem we tackled, recognizing the function in terms of exponents was the first step. Converting \( \sqrt[3]{x^2} \) to \( x^{2/3} \) and \( x\sqrt{x} \) to \( x^{3/2} \) highlights how exponentials arise in algebraic expressions that include roots. By expressing them as powers, it becomes easier to apply rules of differentiation and integration.

• Fractional exponents like \( x^{a/b} \) represent roots, where \( b \) is the root's degree and \( a \) is the regular exponent.

By breaking down expressions into their exponential forms, one can apply the power rule without difficulty, simplifying the process of finding antiderivatives.

Differentiation Verification

After finding an antiderivative, it's essential to confirm its accuracy by differentiating it and checking if it matches the original function. This verification step is crucial in ensuring that no mistakes were made during the integration process.

• When differentiating, apply the power rule again, but this time in reverse: subtract 1 from the exponent and multiply by the original exponent.
• Verify that the result of your differentiation resembles the given original function.

For instance, our antiderivative \( F(x) = \frac{3}{5}x^{5/3} + \frac{2}{5}x^{5/2} + C \) was differentiated back to its original form \( f(x) = x^{2/3} + x^{3/2} \). Differentiating each term accurately retrieves this original function, confirming that the integration was performed correctly.

Engaging in this check-up validates your results, allowing confidence in the solution, and highlighting any computational errors that may have occurred.