Question

Throughout this section we computed several integrals relating to the triangular region $\mathrm{\Omega }$with vertices (1, 1), (2, 0), and (2, 3). In Exercises , you are asked to provide the details of those computations.

Show that the area of $\mathrm{\Omega }$is $\frac{3}{2}$by using the area formula for triangles and by evaluating the integral role="math" localid="1650628428282" ${\int }_1^2 {\int }_{-\mathrm{x}+2}^{2\mathrm{x}-1} \mathrm{dydx}$

Short answer

The triangle of area is$\mathrm{\Omega }=\frac{3}{2}$

Step-by-step solution

Given information 

The given integral is${\int }_1^2 {\int }_{-\mathrm{x}+2}^{2\mathrm{x}-1} \mathrm{dydx}$

Finding the integral value 

The goal of this problem is to demonstrate that the area of is by computing the integral and using the area formula for triangles.

Join the vertices (1,1), (2,0), and (2,3) on a graph

Calculations

Calculate the area of a triangle with three vertices $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right),\text{and}\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$by making use of the formula $\mathrm{\Delta }=\frac{1}{2}\left[{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right]$

Replace the vertices' coordinates

$$\begin{gathered} \mathrm{\Delta }=\frac{1}{2}\left[1\right(0-3)+2(3-1)+2(1-0\left)\right] \\ \mathrm{\Delta }=\frac{{\displaystyle 3}}{{\displaystyle 2}} \end{gathered}$$

Using the integral, get the area of a triangle is $\mathrm{\Omega }={\int }_1^2 {\int }_{-\mathrm{x}+2}^{2\mathrm{x}-1} \mathrm{dydx}$

First, integrate the inner integral $\mathrm{\Omega }={\int }_1^2 \left[{\int }_{-\mathrm{x}+2}^{2\mathrm{x}-1} \mathrm{dy}\right]\mathrm{dx}$

Integrate with relation to y

$\mathrm{\Omega }={\int }_1^2 [\mathrm{y}{]}_{-\mathrm{x}+2}^{2\mathrm{x}-1}\mathrm{dx}$

Substitute the upper and lower bounds

role="math" localid="1650631153817" $\mathrm{\Omega }={\int }_1^2\left[2\mathrm{x}-1-\left(-\mathrm{x}+2\right)\right]\mathrm{dx}$

$\mathrm{\Omega }={\int }_1^2 [3\mathrm{x}-3]\mathrm{dx}$[Demonstrate]

Integrate in relation to x

$\mathrm{\Omega }={\left[\frac{3}{2}{\mathrm{x}}^2-3\mathrm{x}\right]}_1^2$

Substitute the upper and lower bounds

$\begin{array}{r}\mathrm{\Omega }=\left[\frac{3}{2}(2{)}^2-3(2)-\frac{3}{2}(1{)}^2+3\left(1\right)\right]\\ \mathrm{\Omega }=\left[\frac{3}{2}(2{)}^2-3(2)-\frac{3}{2}(1{)}^2+3\left(1\right)\right]\end{array}$

role="math" localid="1650631558065" $\mathrm{\Omega }=\frac{3}{2}$[Establish]

As a result, the triangle's area is $\mathrm{\Omega }=\frac{3}{2}$