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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.PE.37
Chapter 7, Problem 7.PE.37

Evaluate the integrals in Exercises 31–78.
37. ∫(from -1 to 1)dx/(3x-4)

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1
Identify the integral to be evaluated: \(\int_{-1}^{1} \frac{dx}{3x - 4}\).
Check the integrand for any discontinuities or points where the denominator is zero within the interval \([-1, 1]\). Solve \(3x - 4 = 0\) to find such points.
If the integrand is continuous on the interval, proceed to find the antiderivative. Use the substitution method: let \(u = 3x - 4\), then \(du = 3 \, dx\), so \(dx = \frac{du}{3}\).
Rewrite the integral in terms of \(u\): \(\int \frac{dx}{3x - 4} = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{du}{u}\).
Integrate \(\frac{1}{u}\) to get \(\ln|u|\), then substitute back \(u = 3x - 4\). Finally, evaluate the definite integral by applying the limits \(x = -1\) and \(x = 1\) to the antiderivative expression.

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Key Concepts

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

Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating a definite integral involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.
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Antiderivative of Rational Functions

Finding the antiderivative of a rational function like 1/(3x - 4) typically involves recognizing it as a linear denominator. The integral can be solved using substitution or by recalling that ∫1/(ax + b) dx = (1/a) ln|ax + b| + C, where a and b are constants.
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Domain and Discontinuities in Integration

When integrating functions with denominators, it is crucial to check for points where the function is undefined (discontinuities). If the discontinuity lies within the integration limits, the integral may be improper and require special techniques or consideration of limits.
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