Find the total area between y=x+12 and Y=3x+1 between x=0 and x=2.
Hi!!
What do you have so far?
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 :)
To find the total area between the two curves, we need to calculate the areas of the individual regions bounded by the curves and the x-axis. Let's break down the problem into two parts:
1. The region bounded by the curves y = x + 12 and y = 3x + 1 between x = 0 and x = 2.
2. The region below the curve y = x + 12 and above the curve y = 3x + 1 between x = 0 and x = 2.
Part 1: The region bounded by the curves y = x + 12 and y = 3x + 1 between x = 0 and x = 2.
To find the area of this region, we need to calculate the definite integral of the difference of the two functions between the given boundaries:
A₁ = ∫[0, 2] (3x + 1 - (x + 12)) dx
Simplifying the expression inside the integral:
A₁ = ∫[0, 2] (2x - 11) dx
Integrating:
A₁ = [(x^2 - 11x)] evaluated from 0 to 2
Evaluating at the upper limit:
A₁ = (2^2 - 11*2) - (0^2 - 11*0)
A₁ = (4 - 22) - (0 - 0)
A₁ = -18
Since area cannot be negative, we take the absolute value:
A₁ = |-18| = 18
Therefore, the area of the region bounded by y = x + 12 and y = 3x + 1 between x = 0 and x = 2 is 18 square units.
Part 2: The region below the curve y = x + 12 and above the curve y = 3x + 1 between x = 0 and x = 2.
To find the area of this region, we need to calculate the definite integral of the difference of the two functions between the given boundaries:
A₂ = ∫[0, 2] ((x + 12) - (3x + 1)) dx
Simplifying the expression inside the integral:
A₂ = ∫[0, 2] (-2x + 11) dx
Integrating:
A₂ = [(-x^2 + 11x)] evaluated from 0 to 2
Evaluating at the upper limit:
A₂ = (-(2^2) + 11*2) - ((0^2) + 11*0)
A₂ = (-4 + 22) - (0)
A₂ = 18
Therefore, the area of the region below y = x + 12 and above y = 3x + 1 between x = 0 and x = 2 is 18 square units.
To find the total area between the curves, we add the areas of the two regions:
Total Area = A₁ + A₂
= 18 + 18
= 36 square units.
Therefore, the total area between the curves y = x + 12 and y = 3x + 1 between x = 0 and x = 2 is 36 square units.