The total area enclosed by the graphs of
y=10x^2–x^3+x
y=x^2+19x
This is a really long problem I keep getting the answer of 20.25 but it is incorrect I don't know where I am going wrong?
Thanks I see where I made the mistake Thank you!
To find the total area enclosed by the graphs of the given equations, you need to find the points of intersection between the two curves and then calculate the definite integral between those points.
Let's find the points of intersection first by setting the equations equal to each other:
10x^2 – x^3 + x = x^2 + 19x
Rearranging the equation:
-x^3 + 9x^2 + 18x = 0
Factoring out an x:
x(x^2 - 9x - 18) = 0
x = 0 or x^2 - 9x - 18 = 0
Using the quadratic formula, we can solve for x^2 - 9x - 18 = 0:
x = (9 ± sqrt(9^2 - 4*(-18)))/2
x = (9 ± sqrt(225))/2
x = (9 ± 15)/2
For x = (9 + 15)/2 = 12, we have a point of intersection.
For x = (9 - 15)/2 = -3, we have another point of intersection.
To find the total area, we can integrate the difference between the two curves from -3 to 12:
Area = ∫[from -3 to 12] [(x^2 + 19x) - (10x^2 - x^3 + x)] dx
Simplifying the equation inside the integral:
Area = ∫[from -3 to 12] (x^3 - 9x^2 + 18x) dx
Using the power rule, we can integrate each term:
Area = [1/4 x^4 - 3 x^3 + 9 x^2] [from -3 to 12]
Evaluating the integral at the upper and lower limits:
Area = [(1/4 * 12^4 - 3 * 12^3 + 9 * 12^2) - (1/4 * (-3)^4 - 3 * (-3)^3 + 9 * (-3)^2)]
Area = (20736 - 3888 + 324) - (81 + 81 + 27)
Area = 20,976 - 189
Area = 20,787
Therefore, the correct total area enclosed by the graphs of the given equations is 20,787.
To find the area enclosed by the graphs of the given equations, we need to determine the points of intersection between the two graphs first. Then, we can integrate the difference between the two functions over the interval between these points to find the area.
1. To find the intersection points, we need to set the two equations equal to each other:
10x^2–x^3+x = x^2 + 19x
2. Simplify the equation by bringing all the terms to one side:
10x^2–x^3+x - (x^2 + 19x) = 0
Rearranging and combining like terms:
-x^3 + 9x^2 - 18x = 0
3. Factor out an x:
x(-x^2 + 9x - 18) = 0
4. We have one solution x = 0, which can be easily determined from inspection. Now we need to solve the quadratic equation:
-x^2 + 9x - 18 = 0
5. Factor the quadratic equation:
(x - 6)(-x + 3) = 0
So, we have two additional intersection points: x = 6 and x = 3.
Now, let's check the behavior of the two functions to determine which one is on top. We can calculate their values at a single point within each interval (e.g., x = 1, which is between 0 and 3, and x = 4, which is between 3 and 6).
For y = 10x^2 – x^3 + x:
At x = 1: y = 10(1)^2 – (1)^3 + 1 = 10 - 1 + 1 = 10
For y = x^2 + 19x:
At x = 1: y = (1)^2 + 19(1) = 1 + 19 = 20
Therefore, in the interval between x = 0 and x = 3, the graph of y = x^2 + 19x is always above the graph of y = 10x^2 – x^3 + x.
Now we can find the area enclosed between the curves by evaluating the integral of the difference of the two functions over the interval [0, 3]:
∫[0, 3] (x^2 + 19x - 10x^2 + x^3 - x) dx
Simplifying the integrand:
∫[0, 3] (x^3 - 9x^2 + 20x) dx
Integrating term by term:
(1/4)x^4 - (3/4)x^3 + 10x^2 |[0, 3]
Evaluating the definite integral:
[(1/4)(3)^4 - (3/4)(3)^3 + 10(3)^2] - [(1/4)(0)^4 - (3/4)(0)^3 + 10(0)^2]
Simplifying further:
[(1/4)(81) - (3/4)(27) + 90] - [(0 - 0 + 0] = 120.75 - 0 = 120.75
Therefore, the correct answer for the total area enclosed by the graphs of the two equations is 120.75 square units.
Did you get the correct intersections of the two curves?
I had them intersect at x = 0, 3, and 6
between 0 and 3 the parabola is above the curve, so the height is
x^2 + 19x - (10x^2 - x^3 + x) = x^3 - 9x^2 + 18x
between 3 and 6 the cubic lies above the parabola and its height is
-x^3 + 9x^2 - 18x
So you have to take two integrals.
Try it again.