Question

Use calculus to find the area A of the triangle with the given vertices. (0, 0) , (5, 6) , (1, 8)

Answer 1

\[\frac{ 6-0 }{ 5-0 } = \frac{ 6 }{ 5 }x\]
\[\frac{ 8-0 }{ 1-0 }=8x\]
\[\frac{ 8-6 }{ 1-5 }= \frac{ -1 }{ 2 }= \frac{ -1 }{ 2 }x + \frac{ 17 }{ 2 }\]



\[A = \int\limits_{0}^{1}8x-\frac{ 6 }{ 5 }xdx = \int\limits_{0}^{1}\frac{ 34 }{ 5 }x dx= [ \frac{ 34 }{ 5 }x^2]_{0}^{1}=\frac{ 17 }{ 5 }\]
\[A = \int\limits_{1}^{5}[(\frac{ -1 }{ 2 }x+\frac{ 17 }{ 2 })-\frac{ 6 }{ 5 }x]dx = \int\limits_{1}^{5}(\frac{ -5 }{ 10 }x-\frac{ 12 }{ 10 }x)+\frac{ 17 }{ 2 }dx\]
\[=\int\limits_{1}^{5}(\frac{ -17 }{ 10 }x+\frac{ 17 }{ 2 })dx = [\frac{ -17 }{ 20 }x^2+\frac{ 17 }{ 2 }x]_{1}^{5}\]
\[= [(-425/20)+(85/2)]-[(-17/20)+(17/2)]\]
After simplifying this, I would get some big fraction as a wrong answer.