Question

Evaluate the integrals ${\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{0}}^{\mathbf{1}}{\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{x}}^{{\mathbf{e}}^{\mathbf{x}}}\mathit{y}\mathit{d}\mathit{y}\mathit{d}\mathit{x}$ .

Short answer

The value of given integral is ${\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydxis\frac{3e^2-5}{12}$.

Step-by-step solution

Given data

The given equation is${\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx$ .

Concept of Partial differential equation

Integration is a technique of finding a function$\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}$the derivative of which,$\mathit{D}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ , is equal to a given function$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ .

This is indicated by the integral sign "$\mathbf{\int }$ , as in$\mathbf{\int }\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ , usually called the indefinite integral of the function.

Differentiate the equation

Consider the equation as follows:

$$\begin{gathered} {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx={\int }_{x=0}^{1}dx\left({\int }_{y=x}^{e^x}ydy\right) \\ {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx={\int }_{x=0}^{1}dx{\left(\frac{y^2}{2}\right)}_{y=x}^{e^x} \\ {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx=\frac{1}{2}{\int }_{x=0}^1\left(e^{2x}-x^2\right)dx \\ {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx=\frac{1}{2}{\left(\frac{e^{2x}}{2}-\frac{x^3}{3}\right)}_0^1 \end{gathered}$$

Further, solve the equation as follows:

$$\begin{gathered} {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx=\frac{1}{4}\left(e^2-1\right)-\frac{1}{6} \\ {\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx=\frac{3e^2-5}{12} \end{gathered}$$

Therefore, the value of given integral is${\int }_{x=0}^1{\int }_{y=x}^{e^x}ydydx=\frac{3e^2-5}{12}$ .