Problem 5
Question
Sketch the described regions of integration. $$0 \leq x \leq 1, \quad e^{x} \leq y \leq e$$
Step-by-Step Solution
Verified Answer
Shade the region where \( 0 \leq x \leq 1 \) and \( e^x \leq y \leq e \).
1Step 1: Understand the region of integration
We have two inequalities to describe the region in the xy-plane: one for x, which is that it ranges from 0 to 1, and one for y, which ranges from \( e^x \) to \( e \). The first inequality specifies the x-boundaries and the second specifies the y-boundaries.
2Step 2: Identify x-boundaries
The inequality \( 0 \leq x \leq 1 \) tells us that x starts at 0 and ends at 1. This describes a vertical slice of the plane that is one unit wide.
3Step 3: Determine y-boundaries
For each x in the range from 0 to 1, y ranges from \( e^x \) to \( e \). This means the lower boundary is the curve \( y = e^x \) and the upper boundary is the horizontal line \( y = e \).
4Step 4: Sketch the boundaries
Draw the line \( x = 0 \) and \( x = 1 \) on the xy-plane. Then sketch the curve \( y = e^x \) starting at the point (0,1) rising as x increases, and the horizontal line \( y = e \), which intersects the y-axis at a height of approximately 2.718.
5Step 5: Highlight the region
Shade the region between x=0 and x=1, and between the curve \( y = e^x \) and the line \( y = e \). This is the area of integration as defined by the given boundaries.
Key Concepts
InequalitiesSketching RegionsIntegration Boundaries
Inequalities
In calculus, inequalities play a crucial role in defining a region of integration on a graph. They are used to specify the ranges of variables, usually x and y, that form the boundaries of an area. For instance, if given the inequalities \(0 \leq x \leq 1\) and \(e^x \leq y \leq e\), these expressions set constraints for x and y, creating a confined area on the xy-plane.
- The inequality \(0 \leq x \leq 1\) means that x can take any value between 0 and 1, inclusive.
- The inequality \(e^x \leq y \leq e\) means that for each value of x, y can fluctuate from the curve \(y = e^x\) up to the horizontal line \(y = e\).
Sketching Regions
When sketching regions of integration, it’s important to plot the graph accurately, respecting the boundaries set by the inequalities. Begin by plotting any curves and lines described by the equations in the inequalities. Here’s a breakdown to help visualize:
- Draw vertical lines at \(x = 0\) and \(x = 1\) to indicate where x is constrained.
- Sketch the curve \(y = e^x\), starting from the point (0,1). This curve rises smoothly as x increases.
- Plot the horizontal line \(y = e\), which typically intersects the y-axis at approximately 2.718.
Integration Boundaries
Integration boundaries define the limits for evaluating definite integrals. In a region of integration, boundaries like \(0 \leq x \leq 1\) and \(e^x \leq y \leq e\) determine where the integration begins and ends for each variable.
- The intervals \(0 \leq x \leq 1\) set the limits for the variable x. This means when performing integration with respect to x, these are the points where the integration starts and stops.
- Every slice of x from 0 to 1 has a y-bound that ranges between \(y = e^x\) and \(y = e\). These are the limits of integration for y for any given fixed x.
Other exercises in this chapter
Problem 5
Let \(D\) be the region bounded by the paraboloids \(z=8-x^{2}-y^{2}\) and \(z=x^{2}+y^{2} .\) Write six different triple iterated integrals for the volume of \
View solution Problem 5
Sketch the region bounded by the given lines and curves. Then express the region's area as an iterated double integral and evaluate the integral. The curve \(y=
View solution Problem 5
Evaluate the iterated integral. $$\int_{0}^{3} \int_{0}^{2}\left(4-y^{2}\right) d y d x$$
View solution Problem 6
Evaluate the cylindrical coordinate integrals. $$\int_{0}^{2 \pi} \int_{0}^{1} \int_{-1 / 2}^{1 / 2}\left(r^{2} \sin ^{2} \theta+z^{2}\right) d z r d r d \theta
View solution