Question:
Find the area between the curves x+y=5 and x+2y=6.They are both straight lines.
Solution:
I have drawn a graph of the two curves. The red line represents the line x+y=5 and the blue line
represents the line x+2y=6.
The given equations are x+y=5 and x+2y=6.
The first equation is x+y=5.It can be rewritten as y=5-x.
The first function f(x)=y=5-x
The second equation is x+2y=6.It can be rewritten as y=((6-x)/2)
The second function g(x)=y=
6−𝑥
2
Subtracting the first equation from the second equation we get, y=1
Substituting in the first equation we get x=4.
The lines intersect at the point P(4,1).
𝑥2
According to Calculus, Area of the region between the two curves is A=∫𝑥1 [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥
The value of x ranges from 0 to 4.
x1=0 and x2=4 for the integral limit values.
4
4
6−𝑥
)]𝑑𝑥=∫0 [5 −
2
Plugging in the values, we get A=∫0 [5 − 𝑥 − (
𝑥^2
)]
4
=[2x-(
𝑥
from 0 to 4 {By using the properties of definite integrals}
Plugging in the values,
16−0
)
4
=2(4-0)-(
=8-4
=4 square units.
This region lies above the x-axis in the first quadrant in the cartesian plane.
Area between the two curves=4 square units.
4
𝑥
𝑥 − 3 + (2)]dx=∫0 (2 − (2))𝑑𝑥