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Chapter 11: Problem 3

Find \(\int_{0}^{1} \int_{9}^{13}(x+\ln y) d y d x\)

Short Answer

Expert verified
The short answer is \(-2 + 13\ln 13 - 9\ln 9\).

Step by step solution

01

Integrate with respect to y

We start by integrating the given function with respect to \(y\): $$\int (x+\ln y) \, dy$$ We can separate the integral as a sum of integrals: $$\int x \, dy + \int\ln y \, dy$$ Now, we will find the antiderivatives with respect to \(y\). As \(x\) is a constant in the integration with respect to \(y\): $$x\int dy + \int \ln y \, dy$$ The antiderivative of \(\ln y\) with respect to \(y\) is \(y(\ln y - 1)\), so we have: $$x\int dy + y(\ln y - 1)$$ Now, we need to evaluate the antiderivative over the given interval: \(y=(9,13)\):
02

Evaluate the antiderivative between 9 and 13

Now we subtract our function evaluated at the limits given: $$\left[x\int dy + y(\ln y - 1)\right]_{9}^{13} = \left[x(13-9)+((13\ln13-13)-(9\ln9-9))\right] \\ = 4x+(13\ln13-13-9\ln9+9)$$ Our next step is to integrate our function with respect to \(x\) over the interval \((0,1)\):
03

Integrate with respect to x

We have our function, which is a sum of two functions: $$\int_{0}^{1} \left(4x+(13\ln13-13-9\ln9+9) \, dx\right)$$ We can separate it as: $$\int_{0}^{1} 4x \, dx + \int_{0}^{1}(13\ln13-13-9\ln9+9) \, dx$$ Now, we find the antiderivatives of these functions: $$\left[2x^2\right]_{0}^{1} + \left[(13\ln13-13-9\ln9+9)x\right]_{0}^{1}$$
04

Evaluate the antiderivative between 0 and 1

$$\left[2x^2\right]_{0}^{1} + \left[(13\ln13-13-9\ln9+9)x\right]_{0}^{1} = 2(1)^2-2(0)^2+(13\ln13-13-9\ln9+9)(1)-(13\ln13-13-9\ln9+9)(0)$$ Finally, we have: $$2 + 13\ln13 - 13 - 9\ln9 + 9 = 2 - 4 + 13\ln 13 - 9\ln 9 = \boxed{-2 + 13\ln 13 - 9\ln 9}$$

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Key Concepts

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

Definite Integrals
Definite integrals are foundational in calculus and are used to compute the net area under a curve between two specified points. To calculate a definite integral, you need to evaluate the antiderivative of the function at the upper limit of integration and subtract the antiderivative evaluated at the lower limit.

In the textbook exercise provided, the computation of a definite integral begins by integrating with respect to y for a given range, which in this case, is from 9 to 13. After finding the antiderivative, you plug in the limits into this new function to find the accumulated area with respect to y.

A point to emphasize, which is critical in understanding and solving definite integrals, is the concept of the Fundamental Theorem of Calculus. It connects the process of differentiation and integration, stating that the definite integral of a function over an interval is the difference of its antiderivative evaluated at the interval's endpoints.
Antiderivatives
Antiderivatives, also known as indefinite integrals, refer to finding a function whose derivative is the given function. They are a critical component of integration, as seen in our example where the antiderivative of x with respect to y is simply xy, while the antiderivative of ln y is y(ln y - 1).

Quick tips for finding antiderivatives successfully include remembering standard integration formulas and understanding constants' behavior during integration. For instance, any constant multiplied by a variable can be brought out of the integral, simplifying the process.

It's important to practice various integration techniques to become more confident when encountering different types of functions and their respective antiderivatives.
Multiple Integrals Calculus
Multiple integrals extend the concept of single integrals to functions of two or more variables. Double integrals, as observed in the provided example, involve integrating a function over a two-dimensional area.

In our specific exercise, after integrating the function with respect to y, the next step is to integrate with respect to x over the interval from 0 to 1. This layered integration showcases the essence of multiple integrals, where one deals with one variable at a time while considering the others as constants.

Multiple integrals are widely applied in physics and engineering for calculating volumes, mass, and other properties that require integration over several dimensions.
Integration Techniques
Effective integration techniques are necessary to solve complex integrals. The exercise illustrates the use of basic techniques such as linear integration, where the integral of a sum can be treated as the sum of integrals. Additionally, being able to recognize when to use substitution or integration by parts is another essential skill.

In the exercise, we see that the integral is split into two simpler integrals. One integral contains the variable x, and the other is purely a constant with respect to x. Knowing that the integral of a constant times x is straightforward helps streamline the process. Moreover, studying and practicing these techniques will significantly improve one's ability to tackle more intricate integration problems.

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Most popular questions from this chapter

Decide if the points given in polar coordinates are the same. If they are the same, enter \(T\). If they are different, enter \(F\). $$ \begin{array}{l} \text { a.) }\left(5, \frac{\pi}{3}\right),\left(-5, \frac{-\pi}{3}\right) \\ \text { b.) }\left(2, \frac{35 \pi}{4}\right),\left(2,-\frac{35 \pi}{4}\right) \\\ \text { c.) }(0,5 \pi),\left(0, \frac{4 \pi}{4}\right) \\ \text { d.) }\left(1, \frac{141 \pi}{4}\right),\left(-1, \frac{\pi}{4}\right) \\\ \text { e.) }\left(2, \frac{92 \pi}{3}\right),\left(-2, \frac{-\pi}{3}\right) \\\ \text { f. })(5,15 \pi),(-5,15 \pi) \end{array} $$

Let \(x\) denote the time (in minutes) that a person spends waiting in a checkout line at a grocery store and \(y\) the time (in minutes) that it takes to check out. Suppose the joint probability density for \(x\) and \(y\) is $$ f(x, y)=\frac{1}{8} e^{-x / 4-y / 2} $$ a. What is the exact probability that a person spends between 0 to 5 minutes waiting in line, and then 0 to 5 minutes waiting to check out? b. Set up, but do not evaluate, an iterated integral whose value determines the exact probability that a person spends at most 10 minutes total both waiting in line and checking out at this grocery store. c. Set up, but do not evaluate, an iterated integral expression whose value determines the exact probability that a person spends at least 10 minutes total both waiting in line and checking out, but not more than 20 minutes.

For the following two functions \(p(x, y),\) check whether \(p\) is a joint density function. Assume \(p(x, y)=0\) outside the region \(R\). (a) \(p(x, y)=3\), where \(R\) is \(-1 \leq x \leq 1,0 \leq y \leq 0.5\). \(p(x, y) \quad(\square\) is a joint density function \(\quad \square\) is not a joint density function) (b) \(p(x, y)=1,\) where \(R\) is \(1 \leq x \leq 1.5,0 \leq y \leq 2\). \(p(x, y) \quad(\square\) is a joint density function \(\quad \square\) is not a joint density function) Then, for the region \(R\) given by \(-1 \leq x \leq 3,0 \leq y \leq 3,\) what constant function \(p(x, y)\) is a joint density function? \(p(x, y)=\)

Evaluate the double integral \(I=\iint_{\mathbf{D}} x y d A\) where \(\mathbf{D}\) is the triangular region with vertices (0,0),(1,0),(0,6)

Match the following integrals with the verbal descriptions of the solids whose volumes they give. Put the letter of the verbal description to the left of the corresponding integral. (a) \(\int_{0}^{\frac{1}{\sqrt{3}}} \int_{0}^{\frac{1}{2} \sqrt{1-3 y^{2}}} \sqrt{1-4 x^{2}-3 y^{2}} d x d y\) (b) \(\int_{0}^{1} \int_{y^{2}}^{\sqrt{y}} 4 x^{2}+3 y^{2} d x d y\) (c) \(\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} 1-x^{2}-y^{2} d y d x\) (d) \(\int_{0}^{2} \int_{-2}^{2} \sqrt{4-y^{2}} d y d x\) (e) \(\int_{-2}^{2} \int_{4}^{4+\sqrt{4-x^{2}}} 4 x+3 y d y d x\) A. Solid under an elliptic paraboloid and over a planar region bounded by two parabolas. B. Solid under a plane and over one half of a circular disk. C. One half of a cylindrical rod. D. One eighth of an ellipsoid. E. Solid bounded by a circular paraboloid and a plane.
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