Chapter 7: Problem 20
\(\int_{0}^{2} \int_{0}^{x^{2}} e^{y / x} d y d x=\ldots \ldots\)
Short Answer
Expert verified
The value of the integral is \(e^2 - 3\).
Step by step solution
01
Understand the Integral Limits
The given integral is a double integral \[\int_{0}^{2} \int_{0}^{x^{2}} e^{y / x} \, dy \, dx\]where the function \( e^{y/x} \) is to be integrated first with respect to \( y \), from 0 to \( x^2 \), and then with respect to \( x \), from 0 to 2.
02
Perform the Inner Integral
The inner integral: \[\int_{0}^{x^2} e^{y / x} \, dy\]can be solved by substituting \( u = \frac{y}{x} \) which gives \( du = \frac{1}{x} dy \) or \( dy = x \, du \). This transforms the integral into:\[\int_{0}^{x^2/x} e^u \, x \, du = x \int_{0}^{x} e^u \, du\] Integrating \( e^u \) with respect to \( u \) gives:\[x \left[ e^u \right]_0^{x} = x \left( e^x - e^0 \right) = x \left( e^x - 1 \right)\] So, the result of the inner integral is \( x \left( e^x - 1 \right) \).
03
Perform the Outer Integral
Now integrate the result of the inner integral over \( x \):\[\int_{0}^{2} x (e^x - 1) \, dx\]This can be separated into two integrals:\[\int_{0}^{2} x \, e^x \, dx - \int_{0}^{2} x \, dx\]First, consider the integral \( \int x \, e^x \, dx \) using integration by parts where \( u = x \) and \( dv = e^x \, dx \) (thus \( du = dx \) and \( v = e^x \)):\[x \, e^x - \int e^x \, dx = x \, e^x - e^x \]Evaluating from 0 to \( x \) gives:\[\left[ x \, e^x - e^x \right]_0^2 = \left( 2e^2 - e^2 \right) - (0 - e^0) = e^2 - 1\]Now, consider the integral \( \int x \, dx \):\[\frac{x^2}{2} \] Evaluating from 0 to 2 gives:\[\left[ \frac{x^2}{2} \right]_0^2 = \left( 2 \right)\]So the combined evaluation yields:\[ (e^2 - 1) - 2 = e^2 - 3 \]
04
Final Result
The value of the double integral is:\[e^2 - 3\] This is obtained by evaluating the outer integral after solving the inner integral over the given limits.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a crucial technique in calculus for solving integrals that involve products of functions. It's often applied when straightforward integration is challenging.
The formula used is:\[\int u \, dv = uv - \int v \, du\]In this method, you select one part of the integrand as \( u \) (usually the slower growing function, like a polynomial), and another part as \( dv \) (often an exponential or trigonometric function).
In the given exercise, the integration by parts was used to solve the outer integral \( \int x \, e^x \, dx \).
Here, \( u = x \) and \( dv = e^x \, dx \). Therefore, \( du = dx \) and \( v = e^x \).
Plugging these into the integration by parts formula helps break down the problem into more manageable pieces.
The formula used is:\[\int u \, dv = uv - \int v \, du\]In this method, you select one part of the integrand as \( u \) (usually the slower growing function, like a polynomial), and another part as \( dv \) (often an exponential or trigonometric function).
In the given exercise, the integration by parts was used to solve the outer integral \( \int x \, e^x \, dx \).
Here, \( u = x \) and \( dv = e^x \, dx \). Therefore, \( du = dx \) and \( v = e^x \).
Plugging these into the integration by parts formula helps break down the problem into more manageable pieces.
Substitution Method
The substitution method, also known as u-substitution, simplifies the integration process by changing the variable of integration.
It's very similar to reversing the chain rule used in differentiation.In the original problem, substitution was necessary to handle the inner integral \( \int_{0}^{x^2} e^{y/x} \, dy \).
We selected \( u = \frac{y}{x} \), which simplifies the exponent and normalizes the integrand. This choice transforms the differentials as well: \( du = \frac{1}{x} dy \) or \( dy = x \, du \).
This substitution changes the bounds of integration as well, from \( y=0 \) to \( y=x^2 \), the bounds for \( u \) become 0 to \( x \).
Through substitution, the integral becomes easier to solve: \( \int_{0}^{x} e^u \, du \), leading to a result that can be integrated accurately.
It's very similar to reversing the chain rule used in differentiation.In the original problem, substitution was necessary to handle the inner integral \( \int_{0}^{x^2} e^{y/x} \, dy \).
We selected \( u = \frac{y}{x} \), which simplifies the exponent and normalizes the integrand. This choice transforms the differentials as well: \( du = \frac{1}{x} dy \) or \( dy = x \, du \).
This substitution changes the bounds of integration as well, from \( y=0 \) to \( y=x^2 \), the bounds for \( u \) become 0 to \( x \).
Through substitution, the integral becomes easier to solve: \( \int_{0}^{x} e^u \, du \), leading to a result that can be integrated accurately.
Inner Integral
The inner integral refers to the first integration process in a double integral, occurring with the variable closest to the function being integrated.
For this problem, the inner integral is \( \int_{0}^{x^2} e^{y/x} \, dy \), meaning we integrate with respect to \( y \) first.
The task involves simplifying the expression in a way that makes the subsequent "outer" integration manageable. Using substitution makes this possible, transforming the integral into a form where basic integration rules can apply.
After implementing the substitution \( u = \frac{y}{x} \), the expression transforms into \( x \int_{0}^{x} e^u \, du \).
The result of this inner integration is obtained for repeated (or nested) integration later: \( x(e^x - 1) \).
Completing the inner integration is critical for simplifying the complex task of calculating the double integral.
For this problem, the inner integral is \( \int_{0}^{x^2} e^{y/x} \, dy \), meaning we integrate with respect to \( y \) first.
The task involves simplifying the expression in a way that makes the subsequent "outer" integration manageable. Using substitution makes this possible, transforming the integral into a form where basic integration rules can apply.
After implementing the substitution \( u = \frac{y}{x} \), the expression transforms into \( x \int_{0}^{x} e^u \, du \).
The result of this inner integration is obtained for repeated (or nested) integration later: \( x(e^x - 1) \).
Completing the inner integration is critical for simplifying the complex task of calculating the double integral.
Outer Integral
The outer integral is the second phase of the integration process in double integrals. After solving the inner integral, you proceed to integrate the result with respect to the second variable.
In this problem, the outer integral is \( \int_{0}^{2} x(e^x - 1) \, dx \).
The outer integral encapsulates the range \( 0 \) to \( 2 \) for \( x \), and it applies integration by parts to resolve \( x \, e^x \) efficiently.
Firstly, separate the expression into integrable portions: \( \int_{0}^{2} x \, e^x \, dx \) and \( - \int_{0}^{2} x \, dx \).
The resolution of these parts through effective calculus techniques results in the overall solution to the double integral.This step truly wraps up the entire process, leading to the final answer of the problem: \( e^2 - 3 \). Executing the correct techniques in both inner and outer integrals together delivers the complete solution.
In this problem, the outer integral is \( \int_{0}^{2} x(e^x - 1) \, dx \).
The outer integral encapsulates the range \( 0 \) to \( 2 \) for \( x \), and it applies integration by parts to resolve \( x \, e^x \) efficiently.
Firstly, separate the expression into integrable portions: \( \int_{0}^{2} x \, e^x \, dx \) and \( - \int_{0}^{2} x \, dx \).
The resolution of these parts through effective calculus techniques results in the overall solution to the double integral.This step truly wraps up the entire process, leading to the final answer of the problem: \( e^2 - 3 \). Executing the correct techniques in both inner and outer integrals together delivers the complete solution.
