Question

Make the change of variables to evaluate the integral${\mathit{\int }}_{\mathit{0}}^{\mathit{1}}\mathit{d}\mathit{x}{\mathit{\int }}_{\mathit{0}}^{\mathit{x}}\frac{\mathit{(}\mathit{x}\mathit{+}\mathit{y}\mathit{)}{\mathit{e}}^{\mathit{x}\mathit{+}\mathit{y}}}{{\mathit{x}}^{\mathit{2}}}\mathit{d}\mathit{y}$

Short answer

The integral is evaluated as${\int }_0^1\mathrm{dx}{\int }_0^{\mathrm{x}}\frac{(\mathrm{x}+\mathrm{y}){\mathrm{e}}^{\mathrm{x}+\mathrm{y}}}{{\mathrm{x}}^2}\mathrm{dy}={\mathrm{e}}^2-\mathrm{e}-1$ .

Step-by-step solution

Given Condition

The integral given is${\int }_0^1\mathrm{dx}{\int }_0^{\mathrm{x}}\mathrm{dy}\frac{(\mathrm{x}+\mathrm{y}){\mathrm{e}}^{\mathrm{x}+\mathrm{y}}}{{\mathrm{x}}^2}\mathrm{dy}$ .

Concept of integration

Integration is a way of uniting the part to find a whole. Integration is used to define and calculate the area of the region bounded by the graph of functions.

Draw the diagram

 Step 4: Apply the change of variables

The new boundaries of integration will be$u:0\to 1$and$v:0\to 1+u$

First, find$y\left(u,v\right)$and$x\left(u,v\right)$

The values of $y=\frac{vu}{1+u}$and

$x=\frac{v}{1+u}$are known.

 Step 5: Calculate the Jacobian determinant

The Jacobian determinant is found to be the following,

$$\begin{gathered} \left|J\right|=\left|\begin{array}{cc}\frac{\partial x}{\partial v}& \frac{\partial x}{\partial u}\\ \frac{\partial y}{\partial v}& \frac{\partial y}{\partial u}\end{array}\right| \\ =\left|\begin{array}{cc}\frac{1}{1+u}& -\frac{v}{{\left(1+u\right)}^2}\\ \frac{u}{1+u}& \frac{v}{{\left(1+u\right)}^2}\end{array}\right| \\ =\frac{v}{{\left(1+u\right)}^3}+\frac{vu}{{\left(1+u\right)}^3} \\ =\frac{v}{{\left(1+u\right)}^2} \end{gathered}$$

Evaluate the integral

Thus, the integral will be evaluated as follows,

$$\begin{gathered} l={\int }_0^{1}dx{\int }_0^{x}dy\frac{\left(x+y\right)e^{x+y}}{x^2}dy \\ ={\int }_0^{1}du{\int }_0^{1+u}\frac{v{e}^v}{v^2}{\left(1+u\right)}^2\frac{v}{{\left(1+u\right)}^2}dv \\ ={\int }_0^{1}du{\int }_0^{1+u}dv{e}^v \\ ={\int }_0^1\left(e^{1+u}-1\right)du \\ =\left(e^{1+u}-1\right)\overline{){}_0^1} \\ =e^2-e-1 \end{gathered}$$

Hence,${\int }_0^{1}dx{\int }_0^{x}dy\frac{\left(x+y\right)e^{x+y}}{x^2}dy=e^2-e-1$