Question

$\mathbf{\iint }\mathit{d}\mathit{x}\mathbf{}\mathit{d}\mathit{y}$over the area bounded by$\mathit{y}\mathbf{=}\mathbf{Inx}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{e}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{x}$ and the x axis.

Short answer

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

Step-by-step solution

Definition of double integral

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

Drawing the area bounded by the curve

The area is bounded by the graph $y=Inx,y=e+1-x$and x-axis.

Intersecting points identification

From the figure, the two lines $y=Inx,y=e+1-x$intersect at $x=e$.

The integral area splits into two parts. First, have to calculate individual areas then adding both of them gives the total area bounded by the region.

Calculation of the left area bounded by the curve

Calculation of the value, ${\mathrm{I}}_1$:

$$\begin{gathered} {\mathrm{I}}_1={\int }_{\mathrm{x}=1}^{\mathrm{e}}\left[{\int }_{\mathrm{y}=0}^{\mathrm{Inx}}\mathrm{dy}\right]\mathrm{dx} \\ ={\int }_1^{\mathrm{e}}\mathrm{Inxdx} \\ =\mathrm{x}{\overline{)\underline{\mathrm{z}\left(\mathrm{Inx}-1\right)}}}_1^{\mathrm{e}} \\ =1 \end{gathered}$$

Step 5. Calculation of the right side area bounded by the curve

Calculation of the value, $I_2$:

$$\begin{gathered} I_2={\int }_{x=e}^{e+1}\left[{\int }_{y=0}^{e+1-x}dy\right]dx \\ ={\int }_e^{e+1}\left(e+1-x\right)dx \\ ={\overline{)\left(\left(e+1\right)x-\frac{1}{2}{x}^2\right)}}_e^{e+1} \\ =\frac{1}{2} \end{gathered}$$

Total area bounded by the curve

Hence:

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2 \\ =1+\frac{1}{2} \\ =\frac{3}{2} \end{gathered}$$

Therefore, the value is$\frac{3}{2}$.