Chapter 7: Problem 20
Evaluate each iterated integral. \(\int_{-1}^{1} \int_{-1}^{1} x e^{x y} d y d x\)
Short Answer
Expert verified
The iterated integral evaluates to 0.
Step by step solution
01
Understanding the Problem
We need to evaluate an iterated integral of the form \( \int_{-1}^{1} \int_{-1}^{1} x e^{xy} \, dy \, dx \), where the inner integral is with respect to \( y \) and the outer integral is with respect to \( x \).
02
Solve the Inner Integral
First, solve the inner integral \( \int_{-1}^{1} x e^{xy} \, dy \). To solve this, treat \( x \) as a constant and integrate with respect to \( y \): \[ F(y) = \int x e^{xy} \, dy = x \cdot \frac{1}{x} e^{xy} = e^{xy} + C. \]Since we need a definite integral from -1 to 1:\[ \int_{-1}^{1} x e^{xy} \, dy = \left[ e^{xy} \right]_{-1}^{1} = e^{x} - e^{-x}.\]
03
Solve the Outer Integral
Now integrate the result of the inner integral with respect to \( x \):\[ \int_{-1}^{1} \left( e^{x} - e^{-x} \right) \, dx.\]This can be split into two separate integrals:\[\int_{-1}^{1} e^x \, dx - \int_{-1}^{1} e^{-x} \, dx.\]
04
Compute \( \int_{-1}^{1} e^x \, dx \)
Evaluate \( \int_{-1}^{1} e^x \, dx \):\[ \int_{-1}^{1} e^x \, dx = \left[ e^x \right]_{-1}^{1} = e^1 - e^{-1}.\]
05
Compute \( \int_{-1}^{1} e^{-x} \, dx \)
Evaluate \( \int_{-1}^{1} e^{-x} \, dx \):\[ \int_{-1}^{1} e^{-x} \, dx = \left[ -e^{-x} \right]_{-1}^{1} = -e^{-1} - (-e^{1}) = e^{1} - e^{-1}.\]
06
Combine Results from the Outer Integral
Subtract the result of \( \int_{-1}^{1} e^{-x} \, dx \) from \( \int_{-1}^{1} e^x \, dx \):\[ (e^{1} - e^{-1}) - (e^{1} - e^{-1}) = 0.\]
07
Conclude the Calculation
The result of the iterated integral \( \int_{-1}^{1} \int_{-1}^{1} x e^{xy} \, dy \, dx \) is 0.
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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 a crucial concept in calculus. They help us find the total value accumulated by a function over a specific interval. In the iterated integral we are dealing with, the definite integral is computed over two variables: first, with respect to \( y \), then with \( x \). The process involves integrating over a specified range, in this case from -1 to 1, for both variables. This range represents the limits of integration. The evaluation of a definite integral gives us a numerical value, indicating the total accumulated change of the function within those limits.
- Definite integrals have two limits: a lower limit and an upper limit.
- The result of a definite integral is affected by the values at both the lower and upper bounds.
- Definite integrals are geometric interpretations, often representing areas under a curve.
Substitution Integration
Substitution integration is a technique used to simplify the integration process. It often replaces a complex part of an expression with a single variable, making it easier to work with. In the given iterated integral, the substitution is more implicit, as we treat \( x \) as a constant while integrating with respect to \( y \). This allows us to treat the term \( e^{xy} \) more simply and integrate accordingly. Here's how substitution works conceptually:
- Identify a part of the function to substitute with a new variable.
- Adjust the limits of integration if necessary to fit the substitution.
- Integrate the simplified function and then reverse the substitution.
Exponential Functions
Exponential functions are fundamental in calculus due to their unique properties. They are functions of the form \( f(x) = e^{x} \), where \( e \) is a constant approximately equal to 2.718. Exponential functions grow rapidly and possess a special characteristic where their rate of growth is proportional to their value. This property is particularly useful when dealing with integration, as exponential functions have straightforward antiderivatives.
- The derivative and integral of \( e^{x} \) are remarkably similar: \( \frac{d}{dx}e^{x} = e^{x} \) and \( \int e^{x} \, dx = e^{x} + C \).
- When integrating functions involving \( e^{x} \), we often find a straightforward path.
- In iterated integrals, understanding how exponential terms interact with other variables is crucial.