Question

Question: In Problems 7 to 18 evaluate the double integrals over the areas described. To find the limits, sketch the area and compare Figures 2.5 to 2.7.

$\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\left(2x-3y\right)\mathbf{dxdy}$where A is the triangle with vertices (0,0),(2,1),(2,0)

Short answer

The required solution is $\frac{5}{3}$.

Step-by-step solution

Definition of Double Integral

The outer bounds of a double integral must be constant, but the inner limits can be dependent on the outer variable.

Sketch the Integral

The lighter-colored area below is the integration area:

Finding the Integral

Consider the given integral and simplify,

$$\begin{gathered} \int {\int }_{\mathrm{A}}(2\mathrm{x}-3\mathrm{y})\mathrm{dxdy}={\int }_{\mathrm{x}=0}^2\left[{\int }_{\mathrm{y}=0}^{\mathrm{x}/2}(2\mathrm{x}-3\mathrm{y})\mathrm{dy}\right]\mathrm{dx} \\ ={\int }_0^2{\overline{)(2\mathrm{xy}-\frac{3}{2}{\mathrm{y}}^2)}}_0^{\mathrm{x}/2}\mathrm{dx} \\ ={\int }_0^2\left({\mathrm{x}}^2-\frac{3}{8}{\mathrm{x}}^2\right)\mathrm{dx} \\ ={\int }_0^2\frac{5}{8}{\mathrm{x}}^2\mathrm{dx} \\ =\frac{5}{24}{\overline{){\mathrm{x}}^3}}_0^2 \\ =\frac{5\times 8}{24} \\ =\frac{5}{3} \end{gathered}$$

Therefore, the solution is $\frac{5}{3}$.