Question

${\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{1}}^{\mathbf{2}}{\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{x}}^{\mathbf{2}\mathbf{x}}{\mathbf{\int }}_{\mathbf{z}\mathbf{=}\mathbf{0}}^{\mathbf{y}\mathbf{-}\mathbf{x}}\mathit{d}\mathit{z}\mathit{d}\mathit{y}\mathit{d}\mathit{x}$

Short answer

The required integral value is$\frac{7}{6}$ .

Step-by-step solution

Definition of integration

Integration is a way of finding the whole by adding or summing the parts.

Given Information

The integral${\int }_{\mathrm{x}=1}^2{\int }_{\mathrm{y}=\mathrm{x}}^{2\mathrm{x}}{\int }_{\mathrm{z}=0}^{\mathrm{y}-\mathrm{x}}dzdydx$ ,

To find the value for the given integral.

Finding the integral value

Now, integrate the given value,

$$\begin{gathered} \mathrm{I}={\int }_1^2\mathrm{dx}{\int }_{\mathrm{x}}^{2\mathrm{x}}\mathrm{dy}{\int }_0^{\mathrm{y}-\mathrm{x}}\mathrm{dz} \\ ={\int }_1^2\mathrm{dx}{\int }_{\mathrm{x}}^{2\mathrm{x}}\mathrm{dy}(\mathrm{y}-\mathrm{x}) \\ ={\int }_1^2\mathrm{dx}{\left(\frac{1}{2}{\mathrm{y}}^2-\mathrm{xy}\right)}_{\mathrm{x}}^{2\mathrm{x}} \end{gathered}$$

Therefore,

$$\begin{gathered} I={\int }_1^2\left(2x^2-\frac{1}{2}{x}^2-2x^2+x^2\right)dx \\ ={\int }_1^2\frac{1}{2}{x}^{2}dx \\ =\frac{1}{6}{x}^3{\overline{)}}_1^2 \\ =\frac{7}{6} \end{gathered}$$

Therefore, the required solution for the given integral is$\frac{7}{6}$ .