Domanda del libro di testo
Evaluate the integrals in Exercises 33–54.
∫(from ln3 to ln2) (e^x) dx
27
views
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
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 7.2.73
73. Find the area between the curves y=ln(x) and y=ln(2x) from x=1 to x=5.
Guida verificata passo dopo passo1
Identify the two functions given: \( y = \ln(x) \) and \( y = \ln(2x) \). We want to find the area between these curves from \( x = 1 \) to \( x = 5 \).
Determine which function is on top and which is on the bottom in the interval \( [1, 5] \). Since \( \ln(2x) = \ln(2) + \ln(x) \), it is always greater than \( \ln(x) \) for \( x > 0 \). So, \( y = \ln(2x) \) is the upper curve and \( y = \ln(x) \) is the lower curve.
Set up the integral for the area between the curves as \( \int_1^5 [\ln(2x) - \ln(x)] \, dx \).
Simplify the integrand using logarithm properties: \( \ln(2x) - \ln(x) = \ln\left(\frac{2x}{x}\right) = \ln(2) \). So the integral becomes \( \int_1^5 \ln(2) \, dx \).
Evaluate the integral by integrating the constant \( \ln(2) \) over \( [1, 5] \), which is \( \ln(2) \times (5 - 1) \). This gives the area between the curves.
Risposta video verificata per un problema simile:
Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Definite Integrals and Area Between Curves
The area between two curves over an interval is found by integrating the difference of their functions. Specifically, the integral of the upper function minus the lower function from the lower to upper bounds gives the enclosed area.
Video consigliato:
05:23
Finding Area Between Curves on a Given Interval
Properties of Logarithmic Functions
Understanding logarithmic functions, such as ln(x) and ln(2x), is essential. Using log properties like ln(2x) = ln(2) + ln(x) helps simplify the integrand and makes the integration process more straightforward.
Video consigliato:
Percorso guidato
06:21Properties of Functions
Evaluating Definite Integrals Involving Logarithms
Integrating logarithmic functions often requires integration by parts or recognizing standard integral forms. Evaluating the definite integral involves substituting the limits after integration to find the exact area.
Video consigliato:
Percorso guidato
05:43Definition of the Definite Integral
Pratica correlata
Domanda del libro di testo
Evaluate the integrals in Exercises 41–60.
43. ∫6cosh(x/2 - ln3)dx
21
views
Domanda del libro di testo
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
77. y = log₃(((x + 1)/(x − 1))^(ln 3))
28
views
Domanda del libro di testo
128. Derive the formula dy/dx = 1/(1+x²) for the derivative of y = arctan(x) by differentiating both sides of the equivalent equation tan(y)=x.
41
views
Domanda del libro di testo
Evaluate the integrals in Exercises 39–56.
49. ∫3sec²t/(6 + 3tan(t)) dt
20
views
Domanda del libro di testo
In Exercises 73 and 74, repeat the steps above to solve for the functions y=f(x) and x=f^(-1)(y) defined implicitly by the given equations over the interval.
73. y^(1/3) - 1 = (x+2)³, -5 ≤ x ≤ 5, x_0 = -3/2
15
views