91Ó°ÊÓ

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

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

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 $\mathrm{Δ}=\frac{1}{2}\left[{\mathrm{x}}_1\left({\mathrm{y}}_2−{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3−{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1−{\mathrm{y}}_2\right)\right]$

Replace the vertices' coordinates

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

Using the integral, get the area of a triangle is $\mathrm{Î\copyright }={∫}_1^2 {∫}_{−\mathrm{x}+2}^{2\mathrm{x}−1} \mathrm{dydx}$

First, integrate the inner integral $\mathrm{Î\copyright }={∫}_1^2 \left[{∫}_{−\mathrm{x}+2}^{2\mathrm{x}−1} \mathrm{dy}\right]\mathrm{dx}$

Integrate with relation to y

$\mathrm{Î\copyright }={∫}_1^2 [\mathrm{y}{]}_{−\mathrm{x}+2}^{2\mathrm{x}−1}\mathrm{dx}$

Substitute the upper and lower bounds

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

$\mathrm{Î\copyright }={∫}_1^2 [3\mathrm{x}−3]\mathrm{dx}$[Demonstrate]

Integrate in relation to x

$\mathrm{Î\copyright }={\left[\frac{3}{2}{\mathrm{x}}^2−3\mathrm{x}\right]}_1^2$

Substitute the upper and lower bounds

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

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

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