Question

In the problems of this section, set up and evaluate the integrals by hand and check your results by computer.

${\mathbf{\int }}_{\mathbf{x}\mathbf{-}\mathbf{0}}^{\mathbf{4}}{\mathbf{\int }}_{\mathbf{y}\mathbf{-}\mathbf{0}}^{\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{y}\mathbf{-}\mathbf{dy}\mathbf{-}\mathbf{dx}$

Short answer

Value of given integral is $\frac{8}{3}$.

Step-by-step solution

Integrating the integral with respect to y first

Now using the by-parts method of integration for the given integral.

${\int }_{x-0}^4{\int }_{y-0}^{x/2}y-dy-dx={\int }_{x-0}^4\left({\int }_{y-0}^{x/2}y-dy\right)dx$

Using the n-th rule of integralrole="math" localid="1658833631248" $\left(\int {\mathrm{x}}^{\mathrm{n}}-\mathrm{dx}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\right)$.

role="math" localid="1658833730674" ${\int }_{x-0}^4{\int }_{y-0}^{x/2}\mathrm{y}-\mathrm{dy}-\mathrm{dx}={\int }_{\mathrm{x}-0}^4{\left(\frac{1}{2}{\mathrm{y}}^2\right)}_0^{\mathrm{x}/2}-\mathrm{dx}$

Now put the upper and lower limit into the integral and calculate.

role="math" localid="1658833596213" $$\begin{gathered} {\int }_{x-0}^4{\int }_{y-0}^{x/2}y-dy-dx={\int }_{x-0}^4\frac{1}{2}\left({\left(\frac{x}{2}\right)}^2-0^2\right)-dx \\ {\int }_{x-0}^4{\int }_{y-0}^{x/2}y-dy-dx={\int }_{x-0}^4\frac{x^2}{8}dx \end{gathered}$$

Now integrating the integral get in step 1 with respect to x

Integrating with respect to x,

${\int }_{x-0}^4{\int }_{y-0}^{x/2}y-dy-dx={\int }_{x-0}^4\frac{x^2}{8}dx$

Now again using the n-th rule of integral $\left(\int {\mathrm{x}}^{\mathrm{n}}-\mathrm{dx}=\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\right)$.

$$\begin{gathered} {\int }_{x-0}^4{\int }_{y-0}^{x/2}\mathrm{y}-\mathrm{dy}-\mathrm{dx}={\left(\frac{x^3}{8\times 3}\right)}_4^0 \\ {\int }_{x-0}^4{\int }_{y-0}^{x/2}\mathrm{y}-\mathrm{dy}-\mathrm{dx}=\left(\frac{1}{24}\right)\left(4^2-0^3\right) \\ {\int }_{x-0}^4{\int }_{y-0}^{x/2}\mathrm{y}-\mathrm{dy}-\mathrm{dx}=\frac{64}{24} \\ {\int }_{x-0}^4{\int }_{y-0}^{x/2}\mathrm{y}-\mathrm{dy}-\mathrm{dx}=\frac{8}{3} \end{gathered}$$

Therefore, the value of the given integral is $\frac{8}{3}$.