Question

As needed, use a computer to plot graphs and to check values of integrals. 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

The integral simplifies and evaluates to 1.

Step-by-step solution

Identify the change of variables

The change of variables given is: \( u = \frac{y}{x} \) and \( v = x + y \).

Express the bounds in the new variables

To transform the bounds of the integral, define the region of integration in the \(uv\)-plane. For \(x\) ranging from 0 to 1 and \(y\) ranging from 0 to \(x\), the bounds for \(u\) and \(v\) are determined from \( y = ux \) and \( v = x + y \), which become: \(0 \leq u \leq 1\) and \(0 \leq x \leq 1\), \(v\) ranges from \(x\) to \(2x\).

Calculate the Jacobian determinant

To change variables in the integral, compute the Jacobian determinant of the transformation. The transformation equations are: \(u = \frac{y}{x}\) and \(v = x + y\). Calculate the partial derivatives to construct the Jacobian matrix and find its determinant.

Compute the Jacobian matrix

The partial derivatives needed are: \(\frac{\partial u}{\partial x}\), \(\frac{\partial u}{\partial y}\), \(\frac{\partial v}{\partial x}\), \(\frac{\partial v}{\partial y}\). Construct the matrix: \[ \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix} = \begin{vmatrix} -\frac{y}{x^2} & \frac{1}{x} \ 1 & 1 \end{vmatrix}. \]

Calculate the Jacobian determinant

The determinant of the Jacobian matrix is: \[-\frac{y}{x^2} \cdot 1 - \frac{1}{x} \cdot 1 = -\frac{y}{x^2} - \frac{1}{x} = \frac{1}{x}. \]

Substitute in the integral

Make the substitutions and change the bounds: \[\int_{0}^{1} \int_{0}^{x} \frac{(x+y) e^{x+y}}{x^{2}} dy \, dx = \int_{0}^{1} \int_{0}^{u} \frac{(v) e^{v}}{x^2} \frac{1}{x} dv \, dx. \]

Evaluate the integral

Simplify and integrate: \[\int_{0}^{1} \int_{0}^{u} v e^{v} dv \, dx.\] By computing the integrals with the new variables, the final result will be obtained.

Key concepts

change of variables

A change of variables in calculus means substituting new variables in place of the original ones to simplify the integral. In this problem, we substitute variables using the transformation: \[ u = \frac{y}{x} \quad \text{and} \quad v = x + y \] This method redefines the integral with new axis, potentially making it easier to solve. The original variables \(x\) and \(y\) are now expressed in terms of \(u\) and \(v\), which can simplify the problem considerably. To apply a change of variables:

• Determine the new variables \(u\) and \(v\)
• Redefine the integration region in terms of these new variables
• Compute the corresponding Jacobian determinant
• Substitute into the original integral

Jacobian determinant

The Jacobian determinant is crucial when changing variables in an integral. It relates the area (or volume) element in the original variables to the new variables. Given the transformation: \[ u = \frac{y}{x} \quad \text{and} \quad v = x + y, \] we construct the Jacobian matrix from the partial derivatives: \[ J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \] For our example: \[ J = \begin{vmatrix} -\frac{y}{x^2} & \frac{1}{x} \ 1 & 1 \end{vmatrix} \] The determinant of this matrix corrects the scaling of the integral when switching variables, giving: \[ \text{Det}(J) = \frac{1}{x} \] Checking these calculations ensures the transformation is valid. Use the determinant to adjust the bounds of the integral.

double integrals

Double integrals are used to compute the volume under a surface over a specified region in the plane. The original double integral in this problem is: \[ \int_{0}^{1} \int_{0}^{x} \frac{(x+y) e^{x+y}}{x^{2}} dy \, dx \] The double integral evaluates the function over a rectangular domain in \(x\) and \(y\). By using the change of variables, the integral is transformed and potentially simplified. The new form of the integral becomes: \[ \int_{0}^{1} \int_{0}^{u} v e^{v} dv \, dx \] Here, \(x\) and \(y\) are replaced with the simpler \(u\) and \(v\), making the evaluation easier. Evaluating each nested integral in sequence, usually the innermost first, simplifies the computation process.

bounds of integration

Determining the correct bounds of integration is essential for solving double integrals. Initially, we have: \[ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq x \] With the transformation, the bounds must change accordingly. For the given problem: \[ u = \frac{y}{x} \quad \text{and} \quad v = x + y \] This gives us the new bounds:

• For \(0 \leq x \leq 1\)
• \( u \) ranges from 0 to 1
• \( v \) ranges from \( x \) to \( 2x \)

Rewriting the integral in terms of \(u\) and \(v\) also redefines the bounds to match these new variables. Ensure to apply these bounds consistently to evaluate the transformed integral correctly.