Skip to main content
Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 5 - IntegralsProblema 5.PE.17
Capitolo 5, Problema 5.PE.17

Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.
√x + √y = 1, x = 0, y = 0
Graph showing the region bounded by the curve √x + √y = 1 and the x and y axes, shaded in blue.

Guida verificata passo dopo passo
1
Rewrite the given curve equation \(\sqrt{x} + \sqrt{y} = 1\) to express \(y\) in terms of \(x\). Start by isolating \(\sqrt{y}\): \(\sqrt{y} = 1 - \sqrt{x}\).
Square both sides to solve for \(y\): \(y = (1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x\).
Identify the region bounded by the curve and the coordinate axes: \(x=0\), \(y=0\), and the curve \(y = (1 - \sqrt{x})^2\) from \(x=0\) to \(x=1\).
Set up the integral for the area of the region as \(\int_0^1 y \, dx = \int_0^1 (1 - 2\sqrt{x} + x) \, dx\).
Evaluate the integral step-by-step (without calculating the final value): integrate each term separately: \(\int_0^1 1 \, dx\), \(\int_0^1 -2\sqrt{x} \, dx\), and \(\int_0^1 x \, dx\), then sum the results to find the total area.

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:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Area Between Curves

The area between curves is found by integrating the difference between the upper and lower functions over a given interval. When bounded by axes and a curve, the area can be computed by setting up an integral with appropriate limits reflecting the region.
Video consigliato:
05:23
Finding Area Between Curves on a Given Interval

Implicit Functions and Curve Manipulation

The given curve √x + √y = 1 is implicit. To find the area, it is often necessary to express y as a function of x or vice versa. This involves isolating one variable and squaring both sides carefully to avoid extraneous solutions.
Video consigliato:
Percorso guidato
05:14
Finding The Implicit Derivative

Definite Integration with Variable Substitution

Calculating the area under curves involving roots often requires substitution to simplify the integral. For example, substituting u = √x or v = √y can transform the integral into a more manageable form, facilitating evaluation.
Video consigliato:
04:27
Substitution With an Extra Variable
Pratica correlata
Domanda del libro di testo

In Exercises 75–78, find dy/dx.

y = ∫(from x to 1) (6/(3 + t^4))dt

14
views
Domanda del libro di testo

If ∫₀² ƒ(x) dx = π, ∫₀² 7g(x) dx = 7, and ∫₀¹ g(x) dx = 2, find the value of each of the following.

_

d. ∫₀² √2ƒ(x) dx

27
views
Domanda del libro di testo

If ∫²₋₂ 3ƒ(x) dx = 12, ∫⁵₋₂ ƒ(x) dx = 6, and ∫⁵₋₂ g(x) dx = 2, find the value of each of the following.


d. ∫⁵₋₂ (-πg(x)) dx

23
views
Domanda del libro di testo

Definite Integrals


In Exercises 5–8, express each limit as a definite integral. Then evaluate the integral to find the value of the limit. In each case, P is a partition of the given interval, and the numbers cₖ are chosen from the subintervals of P.


     n

 lim  ∑ (cos(cₖ/2)) ∆xₖ, where P is a partition of [-π, 0]

 ∥P∥→0  k = 1

46
views
Domanda del libro di testo

If ∫₀² ƒ(x) dx = π, ∫₀² 7g(x) dx = 7, and ∫₀¹ g(x) dx = 2, find the value of each of the following.


b. ∫₁² g(x) dx

21
views
Domanda del libro di testo

In Exercises 11–14, find the total area of the region between the graph of f and the x-axis.

ƒ(x) = 5 - 5x²/³, -1 ≤ x ≤ 8

32
views