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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.8.6

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₋₈¹ dx / x^(1/3)

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1
Identify the integral to be evaluated: \(\int_{-8}^{1} \frac{dx}{x^{1/3}}\).
Rewrite the integrand using exponent rules: \(\frac{1}{x^{1/3}} = x^{-1/3}\), so the integral becomes \(\int_{-8}^{1} x^{-1/3} \, dx\).
Use the power rule for integration, which states that for \(\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C\), provided \(n \neq -1\). Here, \(n = -\frac{1}{3}\), so \(n + 1 = \frac{2}{3}\).
Apply the power rule to get the antiderivative: \(\int x^{-1/3} \, dx = \frac{x^{2/3}}{2/3} + C = \frac{3}{2} x^{2/3} + C\).
Evaluate the definite integral by substituting the limits: calculate \(\left. \frac{3}{2} x^{2/3} \right|_{-8}^{1}\), which means compute \(\frac{3}{2} (1)^{2/3} - \frac{3}{2} (-8)^{2/3}\).

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Key Concepts

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

Improper Integrals

Improper integrals involve integrands with infinite discontinuities or infinite limits. When the integrand is undefined or unbounded at a point within the interval, the integral is evaluated as a limit approaching that point to determine convergence.
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Integrating power functions of the form x^n involves using the formula ∫x^n dx = (x^(n+1))/(n+1) + C, valid for all n ≠ -1. This rule helps evaluate integrals where the integrand is a power of x, including fractional exponents.
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