Find the total area between y=x+12 and Y=3x+1 between x=0 and x=2.

Answer 1

Hi!!
What do you have so far?

Answer 2

where do they cross?
x + 12 = 3 x + 1
2 x = 11
x = 5.5
so y = x+12 is above y = 3x+1 in the domain of this problem

find area below y = x+12
Now I assume you do not know calculus so will make a quadrilateral
at x = 0, y = 12
at x = 2, y = 14
so at center y = 13
so area between x axis and line is 2*13 = 26
Now find area under y=3x+1 the same way
at x = 0, y = 1
at x = 2, y = 7
so in middle y = 4
area between x axis and line is 2*4 = 8
so
difference = answer = 26-8 = 18
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check using calculus
area = integral (-2x+11)dx from 0 to 2
= -x^2 + 11 x
= -4+22
= 18 sure enough :)