Textbook Question
In Exercises 75–78, find dy/dx.
y = ∫(from x to 1) (6/(3 + t^4))dt
Ch. 5 - Integrals
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- 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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 5, Problem 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
Verified step by step guidance1
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.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
Recommended video:
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.
Recommended video:
05:14Finding 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.
Recommended video:
04:27Substitution With an Extra Variable
Related Practice
Textbook Question
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
Textbook Question
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
1
views
Textbook Question
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
1
views
Textbook Question
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
1
views
Textbook Question
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