Domanda del libro di testo
Evaluate the integrals in Exercises 31–78.
77. ∫dt/((t+1)√(t²+2t-8))
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 7, Problema 7.PE.37
Evaluate the integrals in Exercises 31–78.
37. ∫(from -1 to 1)dx/(3x-4)
Guida verificata passo dopo passo1
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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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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Percorso guidato
05:43
Definition of the Definite Integral
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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Percorso guidato
05:50Antiderivatives
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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5:10Finding the Domain and Range of a Graph
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