Chapter 5: Problem 27

Make the change of variables $u=y / x, v=x+y$, to evaluate the integral $$ \int_{0}^{1} d x \int_{0}^{x} \frac{(x+y) e^{x+y}}{x^{2}} d y $$

Short Answer

Expert verified

The result of the transformed integral is \(\int_0^1 du \frac{1}{(1+u)^2} = checkout x-int.

Step by step solution

Define the new variables

Let’s use the given change of variables. Define the variables as follows: \[ u = \frac{y}{x} \quad \text{and} \quad v = x + y \]

Express the old variables in terms of the new variables

Solve for the old variables $x$ and $y$ in terms of the new variables $u$ and $v$: \[ y = ux \] \[ v = x + y \rightarrow x = \frac{v}{1+u} \rightarrow y = u \frac{v}{1+u} \]

Compute the Jacobian determinant

The Jacobian determinant helps to transform the area elements: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} -\frac{v}{(1+u)^2} & \frac{1}{1+u} \ \frac{v}{(1+u)^2} & \frac{u}{1+u} \end{vmatrix} \rightarrow J = \frac{v}{(1+u)^2} \]

Apply the Jacobian and adjust the integrand

The integrand and differentials transform as follows: \[ \frac{(x+y) \exp(x+y)}{x^2} dx dy = \exp(v) dv du \cdot J \]Given \[ J = \frac{v}{(1+u)^2} \] so: \[ \exp(v) dv du \cdot \frac{v}{(1+u)^2} \]

Define the new limits of integration

Determine how the limits of integration change under the new variables: Original limits: \[ 0 \leq y \leq x 0 \leq x \leq 1 \]New limits:\[\ 0 \leq u < 1 \]\[ 0 \leq v \leq \frac{v}{v+1} \]

Evaluate the double integral

Transform the double integral in new coordinates: \[ \int_0^1 du \int_0^{u/(1+u)} \frac{v e^{v}}{(1+u)^2} \ dv \exp(v) \cdot v = 1 = \frac{1}{(1+u)^2} \frac{v e^{v}}{(1+u)^2} dv du\] Use these new limits of integration and compute the integral.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Change of Variables

When solving multiple integrals, sometimes it's easier to change the variables. This technique allows us to simplify the integrand and limits. In this exercise:

We start with the variables $x$ and $y$. The challenge is to integrate a complex expression over a region.

To make the problem more manageable, we define new variables:

$u = \frac{y}{x}$
$v = x + y$

This change transforms the integration problem into a more straightforward one, where both the integrand and the region's shape might become simpler.

Jacobian Determinant

The Jacobian determinant is vital when changing variables in multiple integrals. It adjusts for the area (or volume) distortion caused by the transformation. Let's break down the process:
After defining the new variables, our next step is to express the old variables in terms of the new ones:

From $u = \frac{y}{x}$ and $v = x + y$, we solve for $x$ and $y$.
Next, we calculate the partial derivatives to form the Jacobian matrix:

\[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \]

This determinant is crucial as it scales the integrand appropriately.

In our exercise, the Jacobian determinant is: $J = \frac{v}{(1+u)^2}$. Applying this, we can properly transform the integral's area elements.

Integration Limits

Changing variables also affects the integration limits. We must convert the original region's bounds to the new variables accurately. Here's a step-by-step guide:

Start with the original limits given in the problem: $0 \leq y \leq x$ and $0 \leq x \leq 1$.
Determine how these limits translate under the new variables $u$ and $v$.

In simple terms:

For $u$: Since $u = \frac{y}{x}$, and $y$ varies from 0 to $x$, $u$ will range from 0 to 1.
For $v$: Since $v = x + y$, and both $x$ and $y$ range between 0 and 1, $v$ will vary within specific bounds which need careful recalculation.

This translates our integral and allows us to evaluate it in a new and simpler coordinate system.

Make the change of variables \(u=y / x, v=x+y\), to evaluate the integral $$ \int_{0}^{1} d x \int_{0}^{x} \frac{(x+y) e^{x+y}}{x^{2}} d y $$