Calculus: Early Transcendentals, Chapter 6, 6.1, Section 6.1, Problem 29

Vertices of the triangle are (0,0) , (3,1) and (1,2)
Equation of the line through (0,0) and (3,1) is,
y-0=((1-0)/(3-0))(x-0)
y=x/3
Equation of the line through (0,0) and(1,2) is,
y-0=((2-0)/(1-0))(x-0)
y=2x
Equation of the line through (3,1) and (1,2) is
y-1=((2-1)/(1-3))(x-3)
y-1=(x-3)/-2
y=-1/2x+3/2+1
y=-1/2x+5/2
Refer to the attached image .Graph of the lines is plotted.
Area of the triangle A =int_0^1(2x-x/3)dx+int_1^3((-x/2+5/2)-x/3)dx
A=int_0^1(5x)/3+int_1^3((-3x-2x)/6+5/2)dx
A=[5/3(x^2/2)]_0^1+int_1^3((-5x)/6+5/2)dx
A=[(5x^2)/6]_0^1+[-5/6(x^2/2)+5/2x]_1^3
A=[5/6(1)^2-5/6(0)^2]+[-5/12(3)^2+5/2(3)]-[-5/6(1)^2+5/2(1)]
A=5/6+(-15/4+15/2)-(-5/12+5/2)
A=5/6+(-15+30)/4-(-5+30)/12
A=5/6+15/4-25/12
A=30/12
A=5/2