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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.2.39

Evaluate the integrals in Exercises 39–56.
39. ∫(from -3 to -2)dx/x

Guida verificata passo dopo passo
1
Identify the integral to be evaluated: \(\int_{-3}^{-2} \frac{1}{x} \, dx\).
Recognize that the integrand \(\frac{1}{x}\) is a standard function whose antiderivative is \(\ln|x|\).
Write the antiderivative: \(F(x) = \ln|x| + C\).
Apply the Fundamental Theorem of Calculus by evaluating \(F(x)\) at the upper and lower limits: compute \(F(-2) - F(-3)\).
Express the definite integral as \(\ln|-2| - \ln|-3|\) and simplify using logarithm properties if needed.

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Improper Integrals

An integral is improper if the integrand is undefined or unbounded within the interval of integration. Since the function 1/x is undefined at x = 0, and the interval [-3, -2] does not include zero, the integral is proper here. However, understanding improper integrals is essential when the interval includes points where the function is undefined.
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Improper Integrals: Infinite Intervals

Definite Integral of 1/x

The integral of 1/x with respect to x is ln|x| + C. When evaluating a definite integral of 1/x over an interval that does not cross zero, you compute the difference of the natural logarithms of the absolute values of the endpoints. This requires careful attention to the domain and sign of x.
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Definition of the Definite Integral

Properties of the Natural Logarithm Function

The natural logarithm function ln(x) is defined for positive x and has properties such as ln(a) - ln(b) = ln(a/b). When integrating 1/x, the absolute value ensures the argument of ln is positive, allowing evaluation over negative intervals by considering ln|x|. Understanding these properties helps correctly evaluate integrals involving 1/x.
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Properties of Functions