Question

${\mathbf{\int }}_{\mathbf{x}\mathbf{=}\mathbf{0}}^{\mathbf{1}}{\mathbf{\int }}_{\mathbf{y}\mathbf{=}\mathbf{0}}^{\mathbf{3}\mathbf{-}\mathbf{3}\mathbf{x}}\mathbf{dydx}$

Short answer

The required solution is$\frac{3}{2}$

Step-by-step solution

Definition of double integral

The double integral of f(x,y) over the area Ain the (x,y) plane as the limit of this sum, and we write it as $\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dxdy}$

Drawing the area bounded by the curve

The area is bounded by the graph y = 3-3x, y-axis and x-axis.

Calculation of the area under the curve in an integral way

At first, the integral is the way it is set up in the problem

$$\begin{gathered} \mathrm{l}={\int }_0^1\mathrm{dx}{\int }_0^{3-3\mathrm{x}}\mathrm{dy} \\ ={\int }_0^1\mathrm{dx}\left(3-3\mathrm{x}\right) \\ ={\overline{)\left(3\mathrm{x}-\frac{3}{2}{\mathrm{x}}^2\right)}}_0^1 \\ =\frac{3}{2} \end{gathered}$$

Calculation of the area under the curve changing boundaries 

The other way, changing the boundaries of the given integration:

$y:0\to 3,\mathrm{and}\mathrm{x}:0\to 1-\frac{\mathrm{y}}{3}$

So, the integration

$$\begin{gathered} l={\int }_0^{3}dy{\int }_0^{1-y/3}dx \\ ={\int }_0^{3}dy\left(1-\frac{y}{3}\right) \\ =3-\frac{9}{6} \\ =\frac{3}{2} \end{gathered}$$

Therefore, both way gives the same results of the integration that is, $\frac{3}{2}$.