91Ó°ÊÓ

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

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$\mathbf{∫}$ , as in$\mathbf{∫}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ , usually called the indefinite integral of the function.

Consider the equation and evaluated as follows:

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

Further, solve the equation as follows:

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

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