Q. 12 - Calculus

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 ${\int }_1^2 {\int }_{-\mathrm{x}+2}^{2\mathrm{x}-1} \mathrm{dydx}$ 

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: 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

Step 3: 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

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

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

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

$\mathrm{\Omega }=\frac{3}{2}$[Establish]

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