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In the following exercises, calculate the integrals by interchanging the order of integration. $$\int_{0}^{2}\left(\int_{0}^{1}\left(x+2 e^{y}-3\right) d x\right) d y$$

Short Answer

Expert verified
The value of the integral is \( 2e^2 - 7 \).

Step by step solution

01

Understand the given integral

The given integral is a double integral: \( \int_{0}^{2}\left(\int_{0}^{1}\left(x+2 e^{y}-3\right) d x\right) d y \). The integration is currently performed with respect to \(x\) first, and then \(y\). We are asked to interchange this order.
02

Interchange the order of integration

Let's interchange the order of integration. This means that we will first integrate with respect to \(y\) and then \(x\). The limits for \(x\) are from 0 to 1, and the limits for \(y\) must be determined accordingly. Since \(y\) originally went from 0 to 2 independently of \(x\), it will remain the same when the order is switched.
03

Set up the new integral

After interchanging the order of integration, the integral can be rewritten as: \( \int_{0}^{1}\left(\int_{0}^{2}\left(x+2 e^{y}-3\right) d y\right) d x \). We will compute the inner integral with respect to \(y\) first.
04

Compute the inner integral with respect to \(y\)

Find the integral \( \int_{0}^{2} (x + 2e^y - 3) \, dy \). Integrate term by term:- \( \int_{0}^{2} x \, dy = x [y]_{0}^{2} = 2x \),- \( \int_{0}^{2} 2e^y \, dy = 2[e^y]_{0}^{2} = 2(e^2 - e^0) = 2(e^2 - 1) \),- \( \int_{0}^{2} 3 \, dy = 3[y]_{0}^{2} = 6 \).Therefore, the result is: \( 2x + 2(e^2 - 1) - 6 \).
05

Simplify the inner integral result

Simplify the expression: \(2x + 2(e^2 - 1) - 6 = 2x + 2e^2 - 2 - 6 = 2x + 2e^2 - 8\).
06

Integrate with respect to \(x\)

Now integrate \( \int_{0}^{1} (2x + 2e^2 - 8) \, dx \).- \( \int_{0}^{1} 2x \, dx = [x^2]_{0}^{1} = 1 \),- \( \int_{0}^{1} 2e^2 \, dx = 2e^2 [x]_{0}^{1} = 2e^2 \),- \( \int_{0}^{1} 8 \, dx = 8[x]_{0}^{1} = 8 \).The final result from integration is: \( 1 + 2e^2 - 8 = 2e^2 - 7 \).
07

Conclusion

After performing all the steps, the result of the integral \( \int_{0}^{2}\left(\int_{0}^{1}\left(x+2 e^{y}-3\right) d x\right) d y \) after interchanging the order of integration is \( 2e^2 - 7 \).

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Key Concepts

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

Order of Integration
In double integrals, the initial order in which integration is performed matters.The original exercise had the order being integrated as follows: first with respect to \( x \), then to \( y \).To interchange the order means to first perform the integration with respect to \( y \) and then to \( x \).This is crucial because it can affect the integration process and sometimes make it simpler.

When changing the order of integration, the function being integrated and the limits of integration may need to be re-evaluated, ensuring they are correctly re-aligned with the new order.This involves re-examining the region of integration, sometimes requiring a sketch to visualize the area more clearly.
Note that the final result of the integral should ultimately remain unchanged, as the value of the definite integral should be independent of the order of integration.
Integration Limits
Identifying the integration limits is essential when solving double integrals.The original limits, \( y \) from 0 to 2 and \( x \) from 0 to 1, were given for the initial setup.

After interchanging the order of integration:
  • The limits for \( x \) remained from 0 to 1 as they were already straightforward.
  • The limits for \( y \), originally independent of \( x \), stayed the same from 0 to 2.
Preserving these limits accurately is crucial, as they define the rectangular region over which the double integral is evaluated.
These limits help determine the span of the integration in their respective directions, effectively bounding the region over which you are integrating.
Inner Integral Calculation
Calculating the inner integral is where you break down the integral into parts, making integration manageable.After you've interchanged the order of integration, you deal with the inner integral first.

In this exercise, the inner integral is conducted with respect to \( y \):\[ \int_{0}^{2} (x + 2e^y - 3) \, dy \]To solve:
  • Integrate each term separately. This decomposes the integral into bite-sized pieces.
  • For \( x \), treat it as a constant: \( x [y]_{0}^{2} \).
  • For \( 2e^y \), apply the exponential rule: \( 2[e^y]_{0}^{2} \).
  • For the constant \( 3 \), use simplicity: \( 3[y]_{0}^{2} \).
This process results in an expression that becomes the subject of the subsequent integral involving \( x \).
Inner integral calculation often involves ensuring that you accurately alter variables and constants to fit the new order of integration.
Integral Simplification
Simplifying the integral once the inner integral has been calculated is the next step.This ensures you deal with an easy-to-integrate expression later on.

From the example, after performing the inner integration, you got:\[ 2x + 2(e^2 - 1) - 6 \]This can further be simplified to:\[ 2x + 2e^2 - 8 \]
This simplification process involves clearing out brackets, combining like terms, and making the expression as succinct as possible before proceeding to the second integration.
It leads to a much more efficient integration process for the final integral, making sure you can focus on simpler mathematical operations in subsequent steps.

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Most popular questions from this chapter

If the charge density at an arbitrary point \((x, y, z)\) of a solid \(E\) is given by the function \(\rho(x, y, z),\) then the total charge inside the solid is defined as the triple integral \(\iiint_{E} \rho(x, y, z) d V,\) Assume that the charge density of the solid \(E\) enclosed by the paraboloids \(x=5-y^{2}-z^{2}\) and \(x=y^{2}+z^{2}-5\) is equal to the distance from an arbitrary point of \(E\) to the origin. Set up the integral that gives the total charge inside the solid \(E\)

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[T] a. Apply the midpoint rule to approximate \(\iiint_{B}e^{-x^{2}} d V\) over the solid \(B=\\{(x, y, z) | 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\}\) by using a partition of eight cubes of equal size. Round your answer to three decimal places. b. Use a CAS to improve the above integral approximation in the case of a partition of \(n^{3}\) cubes of equal size, where \(n=3,4, \ldots, 10\) .

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

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