/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Evaluate each iterated integral.... [FREE SOLUTION] | 91Ó°ÊÓ

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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.
In our exercise, we evaluate \( \int_{-1}^{1} x e^{xy} \, dy \) and \( \int_{-1}^{1} \, dx \), where the precise bounds are critical for the correct calculation, leading to a final result of 0 due to the symmetry and properties of exponential functions.
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.
In our case, recognizing that \( x \) can be treated as a constant simplifies the inner integral to \( e^{xy} \) evaluated as it changes from \( y = -1 \) to \( y = 1 \). Understanding substitution can often lead to finding integrals that seem complicated at first glance.
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.
In the exercise, the term \( x e^{xy} \) demonstrates how exponential functions are featured prominently, necessitating ease of integration when tackling the definite integral over a specified range.

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