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Evaluate the following iterated integrals. $$\int_{0}^{2} \int_{0}^{1} 4 x y d x d y$$

Short Answer

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Question: Evaluate the iterated integral $$\int_{0}^{2}\int_{0}^{1}4xydxdy$$. Answer: 4

Step by step solution

01

Integrate with respect to x

First, we will integrate the function \(4xy\) with respect to x, treating y as a constant: $$\int_{0}^{1} 4 x y d x = 2 x^2 y\Big|_{0}^{1} = 2y$$ Now, we have the single integral with respect to y: $$\int_{0}^{2} 2y dy$$
02

Integrate with respect to y

Next, we will integrate the function \(2y\) with respect to y, using the limits of integration 0 and 2: $$\int_{0}^{2} 2y dy = y^2\Big|_{0}^{2} = 4 - 0 = 4$$ Thus, the value of the iterated integral is 4.

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Key Concepts

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

Integration Techniques
Integration techniques are essential tools in calculus, allowing us to find areas, volumes, and other quantities. One common integration technique is using iterated integrals, especially in multivariable calculus.

In the original exercise, we used a method known as iterated integration. This involves evaluating an integral by breaking it into simpler one-variable integrals. In our case, we start with the inner integral. We first integrate the function \( 4xy \) with respect to \( x \). Here, \( y \) acts as a constant. This is because when integrating with respect to one variable, other variables are treated like constants.

To evaluate \( \int_{0}^{1} 4xy \, dx \), we integrate the expression and then plug in the limits of integration for \( x \). The key is to integrate step by step until the iterated integral is reduced to a simpler form that can be easily evaluated.
Multivariable Calculus
Multivariable calculus extends calculus concepts to more than one variable. It allows us to handle more complex, real-world problems involving multiple changing quantities.

In multivariable calculus, functions often depend on several variables. For example, \( f(x, y) = 4xy \) in our exercise is a function of two variables, \( x \) and \( y \). We use iterated integrals to compute such functions over a given area or volume.
  • First, decide the order of integration. In our case, we integrated first with respect to \( x \), followed by \( y \).
  • This strategy simplifies the problem by reducing it to single-variable integrals, similar to basic calculus.
  • This technique is particularly helpful in physics and engineering, where multiple variables are the norm.
Understanding and mastering multivariable calculus principles allow us to explore more complex systems, providing a framework for tackling problems involving multiple dimensions.
Definite Integrals
Definite integrals are a fundamental part of both single-variable and multivariable calculus. They are used to calculate the accumulation of quantities such as areas under curves.

In the given exercise, the definite integral \( \int_{0}^{2} \int_{0}^{1} 4xy \, dx \, dy \) represents the computation of an area under the surface described by the function \( 4xy \).
  • With definite integration, limits of integration must be specified, such as from 0 to 1 for \( x \), and from 0 to 2 for \( y \).
  • The outcome is a specific numerical value, in this case, 4. This result shows the total value accumulated from integrating the function \( 4xy \) over the given range.
  • Definite integrals play a crucial role in fields like probability, statistics, and any discipline that involves summing continuously varying quantities.
By comprehending definite integrals, we unlock the ability to determine the exact size or value of a function within a set region, providing clarity and precision in practical applications.

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Most popular questions from this chapter

Two integrals to one Draw the regions of integration and write the following integrals as a single iterated integral: $$\int_{0}^{1} \int_{e^{y}}^{e} f(x, y) d x d y+\int_{-1}^{0} \int_{e^{-y}}^{e} f(x, y) d x d y$$

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the cardioid \(r=1+\cos \theta\)

Triangle medians A triangular region has a base that connects the vertices (0,0) and \((b, 0),\) and a third vertex at \((a, h),\) where \(a > 0, b > 0,\) and \(h > 0\) a. Show that the centroid of the triangle is \(\left(\frac{a+b}{3}, \frac{h}{3}\right)\) b. Note that the three medians of a triangle extend from each vertex to the midpoint of the opposite side. Knowing that the medians of a triangle intersect in a point \(M\) and that each median bisects the triangle, conclude that the centroid of the triangle is \(M\)

Choose the best coordinate system and find the volume of the following solid regions. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The wedge cut from the cardioid cylinder \(r=1+\cos \theta\) by the planes \(z=2-x\) and \(z=x-2\)

Evaluate the following integrals using a change of variables of your choice. Sketch the original and new regions of integration, \(R\) and \(S\). \(\iint_{R}\left(\frac{y-x}{y+2 x+1}\right)^{4} d A,\) where \(R\) is the parallelogram bounded by \(y-x=1, y-x=2, y+2 x=0,\) and \(y+2 x=4\)

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