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Reverse the order of integration in the following integrals. $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$

Short Answer

Expert verified
The original integral was given by \(\int_{1}^{e} \int_{0}^{\ln x} f(x, y) dy dx\), and by reversing the order of integration, we obtain the new integral as \(\int_{0}^{1} \int_{e^y}^{e} f(x, y) dx dy\).

Step by step solution

01

Determine the region of integration

Given integral is \(\int_{1}^{e} \int_{0}^{\ln x} f(x, y) dy dx\). The bounds of the integral are: $$ 1 \le x \le e \\ 0 \le y \le \ln x $$ This can be rewritten as inequalities among x and y: $$ x \ge 1 \\ y \le \ln x $$ Based on the given bounds, the region of integration is the area in the xy-plane where x varies from 1 to e and y varies from 0 to ln(x).
02

Find the new limits of integration

To reverse the order of integration, we want to find the new limits of integration for x and y. We do this by expressing x in terms of y: $ x = e^y $ Now, we will determine the minimum and maximum values of x and y. From x = 1 to x = e, y can vary from 0 to 1 (because ln(1) = 0 and ln(e) = 1). Thus, we have the new limits of integration: $$ 0 \le y \le 1 \\ e^y \le x \le e $$
03

Rewrite the integral

Now that we have the new limits of integration, we can rewrite the integral with the new order of integration. The reversed integral is: $$\int_{0}^{1} \int_{e^y}^{e} f(x, y) dx dy$$

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Key Concepts

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

Double Integral
A double integral is a fancy term that essentially means we're integrating a function with two variables. In this case, these variables are usually written as \(x\) and \(y\). Imagine it like a sum, but instead of just summing over one line, we're summing over a whole area. This area is in the \(xy\)-plane.
When dealing with a double integral, it can be shown symbolically like this:
  • \(\int \int f(x, y) \, dx \, dy\)
The function \(f(x, y)\) is the part we're integrating, and the symbols \(dx\) and \(dy\) indicate the areas where we're finding the integral. Double integrals help calculate volumes under surfaces. Now that we understand double integrals involve summing areas, let's see how those areas are defined.
Region of Integration
The region of integration is the area over which we calculate the double integral. Simply put, it tells you where you can evaluate the function.
Given a specified area on the xy-plane, you have conditions or boundaries on both \(x\) and \(y\) values. For example, in this exercise:
  • \(1 \le x \le e\)
  • \(0 \le y \le \ln x\)
This in essence gives us a "box" or "region" in which we are interested. This region is very crucial because it defines the limits - the start and end points - of our integration. Visualize it as fencing an area where all points (\(x, y\)) must lie within.
For the exercise, we found that the region is between \(x\) from 1 to \(e\) and \(y\) from 0 to \(\ln x\). Understanding this region helps in selecting correct integrals.
Limits of Integration
Limits of integration are the bounds that define the region in which the integration occurs. These bounds are where \(x\) and \(y\) start and stop.
In a double integral, we read from the outside integral to the inside. The exercise initially had:
  • Outer integral for \(x\): \(1 \le x \le e\)
  • Inner integral for \(y\): \(0 \le y \le \ln x\)

When we reverse the order of integration, these limits need re-evaluation. This requires expressing \(x\) in terms of \(y\). Here,
  • \(y\) goes from 0 to 1 (\(0 \le y \le 1\))
  • \(x\) varies between \(e^y\) and \(e\) (\(e^y \le x \le e\))
These newly adjusted limits allow us to integrate in the different order \(\int_{0}^{1} \int_{e^y}^{e} f(x, y) \, dx \, dy\). Setting the correct limits is necessary for solving the integral correctly.
Exponential Function
An exponential function is one where a constant base is raised to a variable exponent. Often, we see it represented as \(e^x\), where \(e\) is a special constant approximately equal to 2.71828.
In integrals involving these functions, such as our altered exercise, distinguishing these forms becomes crucial:
  • \(y = \ln x\): The natural log, the inverse of an exponential function, defines one of our region’s boundaries.
  • \(x = e^y\): Describes an exponential relationship where \(x\) is influenced by changes in \(y\).
This reversible property between an exponential and its logarithmic counterpart often simplifies integration processes. Understanding the exponential function is key in setting and re-evaluating the limits when swapping integration orders. Employing properties of exponential functions in integration helps in constructing and simplifying mathematical models in real-life applications.

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. In the iterated integral \(\int_{c}^{d} \int_{a}^{b} f(x, y) d x d y,\) the limits \(a\) and \(b\) must be constants or functions of \(x\) b. In the iterated integral \(\int_{c}^{d} \int_{a}^{b} f(x, y) d x d y,\) the limits \(c\) and \(d\) must be functions of \(y\) c. Changing the order of integration gives \(\int_{0}^{2} \int_{1}^{y} f(x, y) d x d y=\int_{1}^{y} \int_{0}^{2} f(x, y) d y d x\)

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Find the volume of the cap of a sphere of radius \(R\) with height \(h\)

An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. Assume that \(I^{2}=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right)=\) \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y,\) where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi}\) b. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).

The occurrence of random events (such as phone calls or e-mail messages) is often idealized using an exponential distribution. If \(\lambda\) is the average rate of occurrence of such an event, assumed to be constant over time, then the average time between occurrences is \(\lambda^{-1}\) (for example, if phone calls arrive at a rate of \(\lambda=2 /\) min, then the mean time between phone calls is \(\lambda^{-1}=\frac{1}{2}\) min). The exponential distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\)

A mass calculation Suppose the density of a thin plate represented by the region \(R\) is \(\rho(r, \theta)\) (in units of mass per area). The mass of the plate is \(\iint_{R} \rho(r, \theta) d A .\) Find the mass of the thin half annulus \(R=\\{(r, \theta): 1 \leq r \leq 4,0 \leq \theta \leq \pi\\}\) with a density \(\rho(r, \theta)=4+r \sin \theta.\)

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