Question

Question: $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{sinxdxdy}$ where A is the area shown in Figure 2.8

Short answer

The required solution is $\int {\int }_{\mathrm{A}}\mathrm{sinxdxdy}=6$.

Step-by-step solution

Defining the Area of integration.

A definite integral between two points can be used to find the area under a curve between two points.

Separating the area of integration.

Divide the integration region into two rectangles and integrate over each rectangle individually:

$$\begin{gathered} \int {\int }_{\mathrm{A}}\mathrm{sinxdxdy}={\int }_0^{\mathrm{\pi }/2}\mathrm{sinxdx}{\int }_0^2\mathrm{dy}+{\int }_{\mathrm{\pi }/2}^{\mathrm{\pi }}\mathrm{sinxdx}{\int }_0^4\mathrm{dx} \\ =2{\overline{)\left(-\mathrm{cosx}\right)}}_0^{\mathrm{\pi }/2}+4{\overline{)\left(-\mathrm{cosx}\right)}}_{\mathrm{\pi }/2}^{\mathrm{\pi }} \\ =6 \end{gathered}$$

Therefore, the value is $\int {\int }_{\mathrm{A}}\mathrm{sinxdxdy}=6$.