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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 4, Problem 4.7.9a

Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
(2/3)x⁻¹ᐟ³

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1
Identify the function to find the antiderivative of: \(\frac{2}{3} x^{-\frac{1}{3}}\).
Recall the power rule for antiderivatives: for \(x^n\), the antiderivative is \(\frac{x^{n+1}}{n+1} + C\), provided \(n \neq -1\).
Apply the power rule by adding 1 to the exponent: \(-\frac{1}{3} + 1 = \frac{2}{3}\).
Write the antiderivative as \(\frac{2}{3} \cdot \frac{x^{\frac{2}{3}}}{\frac{2}{3}} + C\).
Simplify the expression by canceling the \(\frac{2}{3}\) terms and include the constant of integration \(C\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Antiderivatives (Indefinite Integrals)

An antiderivative of a function is another function whose derivative equals the original function. Finding antiderivatives involves reversing differentiation, often represented as indefinite integrals. For example, the antiderivative of x^n is (x^(n+1))/(n+1) + C, where C is the constant of integration.
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Power Rule for Integration

The power rule for integration states that the integral of x^n with respect to x is (x^(n+1))/(n+1) + C, provided n ≠ -1. This rule is essential for integrating functions with variable exponents, including fractional and negative powers, by increasing the exponent by one and dividing by the new exponent.
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Verification by Differentiation

After finding an antiderivative, verifying it by differentiation ensures correctness. Differentiating the antiderivative should yield the original function. This step confirms that the integration was performed accurately and helps identify any mistakes in the process.
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