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Chapter 7: Problem 6

Evaluate. \(\int_{0}^{1} \int_{-1}^{1}(x+y) d y d x\)

Short Answer

Expert verified
The value of the integral is 1.

Step by step solution

01

- Understand the integral

We need to evaluate the double integral \(\int_{0}^{1} \int_{-1}^{1} (x + y) \, dy \, dx\). This is a double integral over the region where \(x\) ranges from 0 to 1 and \(y\) ranges from -1 to 1.
02

- Integrate with respect to \(y\)

First, hold \(x\) constant and integrate with respect to \(y\). The integral becomes: \(\int_{-1}^{1} (x + y) \, dy\).
03

- Compute the inner integral

Evaluate the inner integral: \(\int_{-1}^{1} (x + y) \, dy = \int_{-1}^{1} x \, dy + \int_{-1}^{1} y \, dy\).
04

- Integrate each term separately

Since \(x\) is constant with respect to \(y\), the terms can be integrated separately: \(x \, \int_{-1}^{1} 1 \, dy + \int_{-1}^{1} y \, dy\).
05

- Evaluate the integrals

Evaluate both integrals: \(\(x \left[ y \right]_{-1}^{1} + \left[ \frac{y^{2}}{2} \right]_{-1}^{1}\).The first integral is \(2x\), and the second integral is 0, since \(\frac{(1)^2}{2} - \frac{(-1)^2}{2} = 0 - 0 = 0\).So, \int_{-1}^{1} (x + y) \, dy = 2x.\)
06

- Integrate with respect to \(x\)

Now, integrate the result with respect to \(x\): \(\int_{0}^{1} 2x \, dx\).
07

- Evaluate the outer integral

Evaluate the outer integral: \(2 \int_{0}^{1} x \, dx = 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1} = 2 \left( \frac{1}{2} - 0 \right) = 1\).

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Key Concepts

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

Iterated Integrals
Iterated integrals allow us to evaluate double integrals by breaking them into a sequence of single integrals.
In this exercise, \( \int_{0}^{1}\ \int_{-1}^{1}(x+y)\ dy\ dx\ \), we first integrate with respect to y, followed by integrating with respect to x.
This step-by-step approach is what defines iterated integrals.
It's helpful for handling more complex regions or functions.
We systematically tackle one variable at a time, simplifying the process.

In essence, iterated integrals convert challenging double integrals into more manageable single integrals, allowing us to evaluate them in a sequential manner.
Limits of Integration
The limits of integration determine the boundaries within which we integrate our function.
For the given exercise, the limits for x are 0 to 1, and for y, they are -1 to 1.
These define the region over which the function \( x + y \) is being integrated.

These limits tell us
  • where to start and stop integrating each variable,
  • ensuring the function is evaluated over the correct range.
The limits of integration must be carefully set to accurately describe the region of interest.
Any mistake in setting these limits can lead to incorrect results.
Integration with Respect to y
When integrating with respect to y, we treat x as a constant.
For the inner integral, \( \int_{-1}^{1}(x+y)\ dy\ \), we hold x fixed and integrate the expression \( x+y \) as a function of y first.

Breaking it down,
  • the integral becomes \( \int_{-1}^{1} x \ dy + \int_{-1}^{1} y \ dy \),
  • where we separately integrate each term.
After calculating
  • the integral of x with respect to y, we find it to be \( 2x \) since it鈥檚 constant over the intervals.
  • The integral of y from -1 to 1 results in 0.
This process results in simplifying the inner integral to 2x.
This step simplifies the complexity of dealing with both variables simultaneously.
Integration with Respect to x
After obtaining the result of the inner integral, we proceed to integrate the expression with respect to x.
Thus, we need to evaluate \( \int_{0}^{1} 2x\ dx \).

Similar to the previous step,
  • we treat 2 as a constant multiplier and focus on integrating x over the range from 0 to 1.
  • The integral of x is \( \frac{x^{2}}{2} \), evaluated between 0 and 1.
When evaluated, this becomes \( 2 \left( \frac{1}{2} - 0 \right) \), which simplifies to 1.

This final step completes the process by providing the value of the original double integral, which is 1.

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Most popular questions from this chapter

In the summer, humidity affects the actual temperature, making a person feel hotter due to a reduced heat loss from the skin caused by higher humidity. The temperature-humidity index, \(T_{h}\), is what the temperature would have to be with no humidity in order to give the same heat effect. One index often used is given by $$ T_{h}=1.98 T-1.09(1-H)(T-58)-56.9 $$ where \(\mathrm{T}\) is the air temperature, in degrees Fahrenheit, and \(H\) is the relative humidity, which is the ratio of the amount of water vapor in the air to the maximum amount of water vapor possible in the air at that temperature. \(H\) is usually expressed as a percentage. Find the temperature- humidity index in each case. Round to the nearest tenth of a degree. $$ T=78^{\circ} \text { and } H=100 \% $$

A one-product company finds that its profit, in millions of dollars, is a function \(P\) given by $$ P(a, n)=-5 a^{2}-3 n^{2}+48 a-4 n+2 a n+300 $$ where \(a\) is the amount spent on advertising, in millions of dollars, and \(n\) is the number of items sold, in thousands. Find the maximum value of \(P\) and the values of \(a\) and \(n\) at which it is attained.

$$ \text { Compute the Jacobian of } f \text { and } g \text { . } $$ $$ f(x, y)=x+3 y, g(x, y)=x-2 y $$

Because wind speed enhances the loss of heat from the skin, we feel colder when there is wind than when there is not. The wind chill temperature is what the temperature would have to be with no wind in order to give the same chilling effect. The wind chill temperature \(W\) is given by \(W(v, T)=\) $$ 91.4-\frac{(10.45+6.68 \sqrt{v}-0.447 v)(457-5 T)}{110}, $$ where \(T\) is the actual temperature as given by a thermometer, in degrees Fahrenheit, and \(v\) is the speed of the wind, in miles per hour. Find the wind chill temperature in each case. Round to the nearest one degree. $$ T=20^{\circ} \mathrm{F}, v=20 \mathrm{mph} $$

Evaluate. Use Simpson's rule with \(n=10\) to approximate the outer integral. \(\int_{1}^{2} \int_{1}^{2} e^{x y} d y d x\)
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