The total area enclosed by the graphs of
y=6x^2–x^3+x and y=x^2+5x is???
Please try yourself first.
sketch a graph.
Find where the two curves intersect.
Then integrate the difference between intersection points.
Note, sketch carefully between x = 0 and x = 5. I think you may find two regions.
Find the points where the curves intersect.
y1 = 6x^2 –x^3 + x = y2 = x^2 + 5x
x^3 - 5x^2 +4x = 0
x(x^2 -5x +4) = 0
x(x-4)(x-1) = 0
One curve (y2) is higher from x=0 to x=1, the other (y1) is higher from x=1 to x=4.
Add the integral from 0 to 1 of (y2 - y1) to the integral from 1 to 4 of (y1 - y2). You need to reverse the sign of the integrand to avoid having one enclosed ares subtracted from the other.
To find the total area enclosed by the graphs of the two given equations, we need to determine the points of intersection between them and then calculate the area between the curves.
Step 1: Find the points of intersection
To find the points where the two curves intersect, we need to set the equations equal to each other:
6x^2–x^3+x = x^2+5x
Rearrange the equation and combine like terms:
6x^2–x^3+x - x^2-5x = 0
Simplify:
5x^2 - x^3 - 4x = 0
Step 2: Factor the equation (if possible)
In this case, we can factor out an x:
x(5x^2 - x^2 - 4) = 0
Factor the quadratic equation:
x(x - 1)(5x + 4) = 0
Set each factor equal to zero and solve for x:
x = 0, x = 1, x = -4/5
These are the x-values where the two curves intersect.
Step 3: Calculate the area between the curves
To find the area between the curves, we integrate the difference of the equations with respect to x over the range of x-values where they intersect.
Area = ∫[(y1 - y2)dx] from x = a to x = b
Let's integrate the difference between the equations:
Area = ∫[(6x^2–x^3+x) - (x^2+5x)] dx
Simplify the integral:
Area = ∫[6x^2 - x^3 + x - x^2 - 5x] dx
Combine like terms:
Area = ∫[-x^3 + 5x^2 - 4x] dx
Now, integrate each term:
Area = [(-1/4)x^4 + (5/3)x^3 - 2x^2] evaluated from x = a to x = b
Plug in the x-values of intersection:
Area = [(-1/4)(b)^4 + (5/3)(b)^3 - 2(b)^2] - [(-1/4)(a)^4 + (5/3)(a)^3 - 2(a)^2]
where a = -4/5 and b = 1.
Substitute the values and calculate the area using the definite integral:
Area = [(-1/4)(1)^4 + (5/3)(1)^3 - 2(1)^2] - [(-1/4)(-4/5)^4 + (5/3)(-4/5)^3 - 2(-4/5)^2]
Now, simplify the expression:
Area = [(-1/4) + (5/3) - 2] - [(-1/4)(256/625) + (5/3)(-64/125) - 2(16/25)]
Area = [3/4 - 2/3 - 8/4] - [(256/1000) + (-320/375) - (32/25)]
Add or subtract each term:
Area = -11/12 - (656/1000 - 320/375 - 32/25)
Area = -11/12 - (-296/1000)
Area = -11/12 + 37/125
Now, find a common denominator:
Area = (-1375/1500) + (444/1500)
Combine the numerators:
Area = -931/1500
Therefore, the total area enclosed by the graphs of the equations y = 6x^2–x^3+x and y = x^2+5x is -931/1500 square units.