Ch. 8 - Integration Techniques

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 8 - Integration Techniques

Problem 8.9.4

Chapter 8, Problem 8.9.4

4. Evaluate ∫ (from 0 to 1) (1/x^(1/5)) dx after writing the integral as a limit.

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Rewrite the integral \( \int_0^1 \frac{1}{x^{1/5}} \, dx \) as \( \int_0^1 x^{-1/5} \, dx \) to express the integrand with a negative exponent, which is easier to integrate.

Recognize that the integrand \( x^{-1/5} \) is undefined at \( x = 0 \) because it involves division by zero, so the integral is an improper integral and must be expressed as a limit.

Express the integral as a limit approaching 0 from the right: \[ \lim_{t \to 0^+} \int_t^1 x^{-1/5} \, dx \]. This handles the improper behavior at the lower limit.

Find the antiderivative of \( x^{-1/5} \) using the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] where \( n = -\frac{1}{5} \). So, \[ \int x^{-1/5} \, dx = \frac{x^{4/5}}{4/5} + C = \frac{5}{4} x^{4/5} + C \].

Evaluate the definite integral by substituting the limits into the antiderivative and then take the limit as \( t \to 0^+ \): \[ \lim_{t \to 0^+} \left( \frac{5}{4} (1)^{4/5} - \frac{5}{4} t^{4/5} \right) \].

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Key Concepts

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

Improper Integrals and Limits

An integral is improper if the integrand is unbounded or the interval is infinite. To evaluate such integrals, rewrite them as limits approaching the problematic point, ensuring the integral converges or diverges.

Power Rule for Integration

The power rule states that ∫ x^n dx = (x^(n+1)) / (n+1) + C for n ≠ -1. This rule helps integrate functions with variable exponents, such as x^(-1/5), by increasing the exponent by one and dividing by the new exponent.

Evaluating Definite Integrals

Definite integrals compute the net area under a curve between two limits. After finding the antiderivative, substitute the upper and lower limits and subtract to find the integral's value.

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