Question

Using integration, find the area of the region bounded by the triangle whose vertices are (-1,2), (1,5) and (3,4).

Answer 1

Hint : Since we need to integrate and find the area, we must find the equation of the 3 sides of the triangle. For finding the equation of the sides use the two point form of the equation of a line. Then integrate along the x axis.

Formula:
Slope: m=$\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Equation of a line when 2 points are given: $(y - {y_{2/1}}) = \left( {x - {x_{2/1}}} \right).m$ , where m is the slope of the line.

Complete step by step solution :
The vertices given are (-1,2), (1,5) and (3,4). Let them be A,B and C respectively.
Finding equation of the line AB-
A=(-1,2) and B=(1,5)
Slope= $\dfrac{{5 - 2}}{{1 - \left( { - 1} \right)}} = \dfrac{3}{2}$
Taking point B in the equation:
$\left( {y - 5} \right) = {\text{ }}\left( {x - 1} \right).\dfrac{3}{2} \\\ 2\left( {y - 5} \right) = 3\left( {x - 1} \right) \\\ 2y - 10 = 3x - 3 \\\ 2y - 3x = 7......\left( i \right) \\\$
Equation of line BC-
B=(1,5) and C=(3,4)
Slope= $\dfrac{{4 - 5}}{{3 - 1}} = \dfrac{{ - 1}}{2}$
Taking point B in the equation:
$\left( {y - 4} \right) = \left( {x - 3} \right)\dfrac{{\left( { - 1} \right)}}{2} \\\ 2\left( {y - 4} \right) = - \left( {x - 3} \right) \\\ 2y - 8 = - x + 3 \\\ 2y + x = 11......\left( {ii} \right) \\\$
Equation of line AC-
A=(-1,2) and C=(3,4)
Slope= $\dfrac{{4 - 2}}{{3 - \left( { - 1} \right)}} = \dfrac{2}{4} = \dfrac{1}{2}$
Taking point C in the equation:
$\left( {y - 4} \right) = \left( {x - 3} \right)\dfrac{1}{2} \\\ 2\left( {y - 4} \right) = \left( {x - 3} \right) \\\ 2y - 8 = x - 3 \\\ 2y - x = 5.....\left( {iii} \right) \\\$

Integrating –
$\int\limits_{ - 1}^1 {AB} + \int\limits_1^3 {BC - } \int\limits_{ - 1}^3 {AC}$ = area of $\Delta ABC$
Since we are integrating along x we find y in terms of x from all the three given equations:
{\text{y = }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3x}}}}{{\text{2}}}{\text{ from}}\left( {\text{i}} \right) \\\ {\text{y = }}\dfrac{{{\text{11}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ from}}\left( {{\text{ii}}} \right) \\\ {\text{y = }}\dfrac{{\text{5}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ from}}\left( {{\text{iii}}} \right) \\\
$= \int\limits_{ - 1}^1 {[\dfrac{{\text{7}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{3x}}}}{{\text{2}}}]{\text{ }} + {\text{ }}} \int\limits_1^3 {[\dfrac{{{\text{11}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{x}}}{{\text{2}}}]{\text{ }} - } {\text{ }}\int\limits_{ - 1}^3 {[\dfrac{{\text{5}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{x}}}{{\text{2}}}} ] \\\ = \int\limits_{ - 1}^1 {\dfrac{{\text{7}}}{{\text{2}}}{\text{ + }}\int\limits_{ - 1}^1 {\dfrac{{{\text{3x}}}}{{\text{2}}}} {\text{ }} + {\text{ }}} \int\limits_1^3 {\dfrac{{{\text{11}}}}{{\text{2}}}{\text{ - }}\int\limits_1^3 {\dfrac{{\text{x}}}{{\text{2}}}} {\text{ }} - } {\text{ }}\int\limits_{ - 1}^3 {\dfrac{{\text{5}}}{{\text{2}}} - \int\limits_{ - 1}^3 {\dfrac{{\text{x}}}{{\text{2}}}} } \\\ = [\dfrac{{7x}}{2}\mathop ]\limits_{ - 1}^1 + [\dfrac{{{\text{3}}{{\text{x}}^2}}}{4}\mathop ]\limits_{ - 1}^1 + [\dfrac{{{\text{11x}}}}{{\text{2}}}\mathop ]\limits_1^3 - [\dfrac{{{{\text{x}}^2}}}{4}\mathop ]\limits_1^3 - [\dfrac{{{\text{5x}}}}{{\text{2}}}\mathop ]\limits_{ - 1}^3 - [\dfrac{{{{\text{x}}^2}}}{4}\mathop ]\limits_{ - 1}^3 \\\ = [\dfrac{{7\left( 1 \right)}}{2} - \dfrac{{7\left( { - 1} \right)}}{2}] + [\dfrac{{{\text{3}}{{\left( 1 \right)}^2}}}{4} - \dfrac{{{\text{3}}{{\left( { - 1} \right)}^2}}}{4}] + \\\ [\dfrac{{{\text{11}}\left( 3 \right)}}{{\text{2}}} - \dfrac{{{\text{11}}\left( 1 \right)}}{{\text{2}}}] - [\dfrac{{{{\left( 3 \right)}^2}}}{4} - \dfrac{{{{\left( 1 \right)}^2}}}{4}] - [\dfrac{{{\text{5}}\left( 3 \right)}}{{\text{2}}} - \dfrac{{{\text{5}}\left( { - 1} \right)}}{{\text{2}}}] - \left[ {\dfrac{{{3^2}}}{4} - \dfrac{{{{\left( { - 1} \right)}^2}}}{4}} \right] \\\ = [\dfrac{7}{2} + \dfrac{7}{2}] + [\dfrac{{\text{3}}}{4} - \dfrac{3}{4}] + [\dfrac{{33}}{{\text{2}}} - \dfrac{{{\text{11}}}}{{\text{2}}}] - [\dfrac{9}{4} - \dfrac{1}{4}] - [\dfrac{{15}}{2} + \dfrac{5}{2}] - [\dfrac{9}{4} - \dfrac{1}{4}] \\\ = 7 + 0 + 11 - 2 - 10 - 2 \\\ = 4{\text{ }} \\\$
The area of the triangle is 4 sq units.

Note : Since we integrated along the x axis the equations were considered in x if we were to integrate along the y axis, the equation would be considered as a whole in y and the limits of integration would be the y coordinates of the triangle.