Question

Evaluate the triple integrals ${\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{1}}^{\mathbf{2}}{\mathbf{\int }}_{\mathbf{z}\mathbf{=}\mathbf{x}}^{\mathbf{2}\mathbf{x}}{\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{0}}^{\mathbf{1}\mathbf{/}\mathbf{z}}\mathit{z}\mathit{d}\mathit{y}\mathit{d}\mathit{z}\mathit{d}\mathit{x}$.

Short answer

The value of given integral ${\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{1/z}zdydzdx$is$\frac{9}{4}$ .

Step-by-step solution

Given data

The given equation is ${\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{1/z}zdydzdx$.

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 and evaluated as follows:

$$\begin{gathered} {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx={\int }_1^{2}dx{\int }_x^{2x}zdz{\int }_0^{\frac{1}{x}}dy \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx={\int }_1^{2}dx{\int }_x^{2x}zdz\left(\frac{1}{x}\right) \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx={\int }_1^2\frac{dx}{x}{\int }_x^{2x}zdz \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx={\int }_1^2\frac{dx}{x}{\left(\frac{1}{2}{z}^2\right)}_x^{2x} \end{gathered}$$

Further, solve the equation as follows:

$$\begin{gathered} {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx=\frac{1}{2}{\int }_1^2\frac{dx}{x}\left(4x^2-x^2\right) \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx=\frac{1}{2}{\int }_1^{2}dx(4x-x) \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx=\frac{3}{4}{\left(x^2\right)}_1^2 \\ {\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx=\frac{9}{4} \end{gathered}$$

Therefore, the evaluation of given integral ${\int }_{x=1}^2{\int }_{z=x}^{2x}{\int }_{y=0}^{\frac{1}{z}}zdydzdx$is$\frac{9}{4}$ .