Chapter 4: Problem 37
In the following exercises, find the average value of the function over the given rectangles. $$ f(x, y)=\sinh x+\sinh y, R=[0,1] \times[0,2] $$
Short Answer
Expert verified
The average value of the function is \( \cosh 1 + \frac{\cosh 2 - 3}{2} \).
Step by step solution
01
Formula for Average Value
The average value of a function over a rectangle is given by \( \frac{1}{\text{Area of } R} \int_{R} f(x, y) \, dA \), where \( R \) is the rectangular region. Hence, the area of \( R \) is \( 2 \times 1 = 2 \).
02
Set up the Integral
To find the integral of \( f(x, y) = \sinh x + \sinh y \) over \( R = [0, 1] \times [0, 2] \), set up the double integral \( \int_{0}^{1} \int_{0}^{2} (\sinh x + \sinh y) \, dy \, dx \).
03
Integrate with Respect to y
Integrate \( \sinh x + \sinh y \) with respect to \( y \):\[ \int_{0}^{2} (\sinh x + \sinh y) \, dy = \left[ 2\sinh x + \cosh y \right]_{0}^{2} = 2\sinh x + \cosh 2 - \cosh 0. \]
04
Evaluate the Bounds for y
Plug in the bounds for \( y \):For \( y = 2 \), \( \cosh 2 \). For \( y = 0 \), \( \cosh 0 \), thus simplifying to: \[ 2\sinh x + (\cosh 2 - 1). \]
05
Integrate with Respect to x
Now integrate the simplified expression \( 2\sinh x + (\cosh 2 - 1) \) with respect to \( x \):\[ \int_{0}^{1} \left( 2\sinh x + (\cosh 2 - 1) \right) \, dx = \left[ 2\cosh x + (\cosh 2 - 1)x \right]_{0}^{1}. \]
06
Evaluate the Bounds for x
Calculate \( 2\cosh 1 - 2 + (\cosh 2 - 1) \times 1 \). Simplify to find:\[ (2\cosh 1 + \cosh 2 - \cosh 2 - 1) = 2(\cosh 1 - 1) + \cosh 2 - 1 = 2\cosh 1 + \cosh 2 - 3. \]
07
Calculate the Average Value
The average value of \( f(x, y) \) over \( R \) is given by:\[ \frac{1}{2} \left( 2\cosh 1 + \cosh 2 - 3 \right) = \cosh 1 + \frac{\cosh 2 - 3}{2}. \]
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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 allows us to integrate over a two-dimensional area. In this exercise, we use a double integral to find the average value of a function over a rectangular region. A double integral is written as \( \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy\, dx \) where \( f(x, y) \) is the function to integrate, \( [a,b] \) is the range for \( x \), and \( [c,d] \) is the range for \( y \).
For the function \( f(x, y) = \sinh x + \sinh y \) given in the exercise, we integrate first with respect to \( y \) from 0 to 2, then with respect to \( x \) from 0 to 1.
For the function \( f(x, y) = \sinh x + \sinh y \) given in the exercise, we integrate first with respect to \( y \) from 0 to 2, then with respect to \( x \) from 0 to 1.
- Integrate \( \sinh x + \sinh y \) with respect to \( y \) over the interval \([0,2]\). This gives the antiderivative \( 2\sinh x + \cosh y \).
- Then, substitute the bounds 2 and 0 for \( y \), which gives us \( 2\sinh x + \cosh 2 - \cosh 0 \).
- Next, the expression \( 2\sinh x + (\cosh 2 - 1) \) is integrated with respect to \( x \) from 0 to 1.
Hyperbolic Functions
Hyperbolic functions like \( \sinh x \) and \( \cosh x \) play a role similar to trigonometric functions but are defined using exponential functions.
When integrating hyperbolic functions:
- The hyperbolic sine function is \( \sinh x = \frac{e^x - e^{-x}}{2} \).
- The hyperbolic cosine function is \( \cosh x = \frac{e^x + e^{-x}}{2} \).
When integrating hyperbolic functions:
- The integral of \( \sinh x \) is \( \cosh x + C \) where \( C \) is the constant of integration.
- Similarly, the integral of \( \cosh x \) results in \( \sinh x + C \).
Rectangular Region Integration
Rectangular region integration involves evaluating a double integral over a specific rectangle defined by its boundaries on the \( x \)- and \( y \)-axes. In this context, the region \( R = [0, 1] \times [0, 2] \) is the rectangle over which the function is averaged.
Understanding the concept of integration over rectangular regions gives you a valuable tool for summarizing how a function behaves over that region.
Understanding the concept of integration over rectangular regions gives you a valuable tool for summarizing how a function behaves over that region.
- The first step is to identify the limits of integration for each variable; \( x \) ranges from 0 to 1, and \( y \) ranges from 0 to 2.
- Set up the double integral by nesting one integral inside the other, choosing which variable to integrate first based on simplicity or order preferences.
- Calculating the integral involves solving the inner integral first and then proceeding to the outer integral.