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91Ó°ÊÓ

Sketch the described regions of integration. $$0 \leq x \leq 1, \quad e^{x} \leq y \leq e$$

Short Answer

Expert verified
Sketch the strip where \(0 \leq x \leq 1\) and the area between \(y = e^x\) and \(y = e\).

Step by step solution

01

Understand the Region Bounds for x

The given range for \(x\) is \(0 \leq x \leq 1\). This tells us that we are considering points between \(x = 0\) and \(x = 1\) on the x-axis. This vertical strip \ will be part of the region of integration.
02

Understand the Region Bounds for y

The inequality \(e^{x} \leq y \leq e\) describes the bounds for \(y\). For a fixed \(x\), \(y\) begins at \(e^x\) and goes up to \(e\). This implies the curve \(y = e^x\) acts as the lower boundary, and \(y = e\) acts as the upper boundary.
03

Sketch the Lower Boundary Curve

Plot the curve \(y = e^x\) for \(0 \leq x \leq 1\). This is an increasing curve starting from the point \((0, 1)\) and moving to \((1, e)\).
04

Plot the Horizontal Upper Boundary

Draw the horizontal line \(y = e\) across the interval \(0 \leq x \leq 1\). This line will cut across the y-axis at the height \(e\), parallel to the x-axis, acting as the upper boundary of the region.
05

Identify and Shade the Region of Integration

The region of integration is bound between the curve \(y = e^x\) and the horizontal line \(y = e\), for \(0 \leq x \leq 1\). Shade the area between these two lines, between \(x = 0\) and \(x = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regions of Integration
In calculus, the region of integration refers to the specific area over which an integration operation takes place. When dealing with double integrals, these regions are typically determined by boundaries set on both the x-axis and the y-axis.

To visualize this, consider a rectangle in the xy-plane. Its sides are defined by the limits on x and y. These limits tell us where we should start and stop integrating.

In the exercise given, the region is defined by the bounds from the step-by-step solution. To recap, our region lies between:
  • The vertical limits of 0 and 1 on the x-axis.
  • The lower curve, which is \(y = e^x\), and the upper line, \(y = e\).
The region of integration here is the area under the exponential curve \(y = e^x\) and capped by the horizontal line \(y = e\), within the vertical strip of \(0 \leq x \leq 1\). This part we sketch out is the complete region of integration for the problem.
Bounds
Bounds in integration essentially provide the limits for the variables you’re working with. They define the perimeter of the region you are integrating over.

In the problem discussed, we see two sets of bounds:
  • For x: The bounds are given as \(0 \leq x \leq 1\). This means that x starts at 0 and ends at 1, forming a vertical strip along the x-axis.
  • For y: The bounds from the exercise say \(e^x \leq y \leq e\). These define the start and stop for y-values at any particular x-value.
Understanding these bounds helps in visualizing the region. They confine the space where all possible points (x, y) must lie to be counted as part of the region of integration.
Graph Sketching
Graph sketching is an important skill that allows you to visually verify the behavior of functions and understand regions of integration more clearly. In this exercise, we are dealing with two particular functions:
  • The curve \(y = e^x\), which is an exponential growth curve.
  • The line \(y = e\), which is horizontal.
To sketch these:Start by plotting the curve \(y = e^x\). Note that when \(x = 0\), \(y = e^0 = 1\), and when \(x = 1\), \(y = e^1 = e\). This curve moves upwards, indicating exponential growth.Next, draw the horizontal line \(y = e\). It cuts through the y-axis at the height dictated by e.Finally, highlight the region of interest. This is the area between the curve and the horizontal line over the interval \(0 \leq x \leq 1\). Shading this area will give a visual confirmation of the region you've identified through your calculations.
Exponential Functions
Exponential functions are a class of mathematical functions commonly written as \(y = a^x\), where \(a\) is a constant greater than one. These functions have distinct characteristics:
  • They always produce positive values for real input x.
  • The function is increasing, meaning as x grows, so does y.
  • The rate of increase is proportional to the function’s current value, resulting in that sharp, upward curve.
In the exercise, the function \(y = e^x\) is our example of an exponential function, where e is Euler's number, approximately 2.718.

Exponential functions like \(y = e^x\) are fundamental in many fields such as finance, biology, and physics due to their properties of constant proportionality—this feature is crucial in modeling growth processes.

Being familiar with these fundamentals makes understanding the broader applications in integrals and other areas much easier.

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Most popular questions from this chapter

Evaluate the integrals. $$\int_{0}^{2} \int_{-\pi}^{0} \int_{\pi / 4}^{\pi / 2} \rho^{3} \sin 2 \phi d \phi d \theta d \rho$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid in the first octant bounded by the coordinate planes, the plane \(x=3,\) and the parabolic cylinder \(z=4-y^{2}\)

Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

A solid right (noncircular) cylinder has its base \(R\) in the \(x y\) -plane and is bounded above by the paraboloid \(z=x^{2}+y^{2} .\) The cylinder's volume is $$V=\int_{0}^{1} \int_{0}^{y}\left(x^{2}+y^{2}\right) d x d y+\int_{1}^{2} \int_{0}^{2-y}\left(x^{2}+y^{2}\right) d x d y$$ Sketch the base region \(R\) and express the cylinder's volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to find the volume.

Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid whose base is the region in the \(x y-\) plane that is bounded by the parabola \(y=4-x^{2}\) and the line \(y=3 x,\) while the top of the solid is bounded by the plane \(z=x+4\)

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