Question

${\mathbf{\int }}_{\mathbf{y}\mathbf{-}\mathbf{1}}^{\mathbf{2}}{\mathbf{\int }}_{\mathbf{x}\mathbf{-}\sqrt{\mathbf{y}}}^{{\mathbf{y}}^{\mathbf{2}}}\mathit{x}\mathit{d}\mathit{x}\mathit{d}\mathit{y}$

Short answer

The required solution is ${\int }_{\mathrm{y}-1}^2{\int }_{\mathrm{x}-\sqrt{\mathrm{y}}}^{{\mathrm{y}}^2}\mathrm{xdxdy}\frac{47}{20}$.

Step-by-step solution

Definition of Definite Integral

The antiderivative and indefinite integrals of a function are closely connected to the definite integral. The indefinite integral, if it exists, has a real numerical value, whereas the latter two represent an unlimited number of functions that differ only by a constant.

Finding the Integral

Consider the given integral,

$$\begin{gathered} {\int }_{\mathrm{y}-1}^2{\int }_{\mathrm{x}-\sqrt{\mathrm{y}}}^{{\mathrm{y}}^2}\mathrm{xdxdy}={\int }_{\mathrm{y}-1}^2\left[\frac{1}{2}{x}^2\overline{){}_{\sqrt{y}}^{y^2}}\right]dy \\ =\frac{1}{2}{\int }_1^2\left(y^4-y\right)dy \\ =\frac{1}{10}{y}^5\overline{){}_1^2-\frac{1}{4}{y}^2}\overline{){}_1^2} \\ =\frac{1}{10}\left(32-1\right)-\frac{1}{4}\left(4-1\right) \\ =\frac{62}{10}-\frac{15}{20} \\ =\frac{47}{20} \end{gathered}$$

Therefore, the solution is ${\int }_{\mathrm{y}-1}^2{\int }_{\mathrm{x}-\sqrt{\mathrm{y}}}^{{\mathrm{y}}^2}\mathrm{xdxdy}=\frac{47}{20}$.