Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 30

Find the area of the triangle with the given vertices $(0,5), (2,-2)$ and $(5,1)$ using Calculus.

We can plot the triangle first to help us evaluate the area.







We can use the point slope form to determine the equations of the line..

at points $(0,5)$ and $(2, -2)$


$
\begin{equation}
\begin{aligned}

y - 5 =& \left( \frac{-2 - 5}{2 - 0} \right) (x - 0 )
\\
\\
y = & - \frac{7}{2} x + 5

\end{aligned}
\end{equation}
$


at points $(0,5)$ and $(5, 1)$


$
\begin{equation}
\begin{aligned}

y - 5 =& \left( \frac{1 - 5}{5 - 0} \right) (x - 0)
\\
\\
y =& - \frac{4}{5} x + 5

\end{aligned}
\end{equation}
$


at points $(5,1)$ and $(2, -2)$


$
\begin{equation}
\begin{aligned}

y - 1 =& \left( \frac{-2 - 1}{2 - 5} \right) (x - 5)
\\
\\
y =& \frac{-3}{-3} (x - 5) + 1
\\
\\
y =& x - 4

\end{aligned}
\end{equation}
$


Now, we can divide the area into two sub region. Let $A_1$ and $A_2$ be the area in the left and right part respectively. So..

By using vertical strip,


$
\begin{equation}
\begin{aligned}

A_1 =& \int^{x_22}_{x_1} (y_{\text{upper}}, y_{\text{lower}}) dx
\\
\\
A_1 =& \int^2_0 \left[ \frac{-4}{5} x + 5 - \left( - \frac{7}{2} x + 5 \right) \right] dx
\\
\\
A_1 =& \left[ \frac{27}{10} \left( \frac{x^2}{2} \right) \right]^2_2
\\
\\
A_1 =& \frac{27}{5} \text{ square units}

\end{aligned}
\end{equation}
$


For the right part,


$
\begin{equation}
\begin{aligned}

A_2 =& \int^5_2 \left[ \frac{-4}{5} x + 5 - (x - 4) \right] dx
\\
\\
A_2 =& \int^{5}_2 \left[ - \frac{9x}{5} + 9 \right] dx
\\
\\
A_2 =& \left[ - \frac{9x^2}{2 (5)} + 9x \right] ^5_2
\\
\\
A_2 =& \frac{81}{10} \text{ square units}

\end{aligned}
\end{equation}
$


Therefore, the total area of the triangle is $\displaystyle A_1 + A_2 = \frac{27}{2}$ square units