3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x.
I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you
To find the area bounded by the curves f(x) = x^3 + x^2 and g(x) = 2x^2 + 2x, you can use definite integrals. The area can be calculated by finding the difference between the integrals of the curves.
First, we need to find the x-values where the two curves intersect. Setting the two equations equal to each other, we have:
x^3 + x^2 = 2x^2 + 2x
Rearranging the equation:
x^3 + x^2 - 2x^2 - 2x = 0
x^3 - x^2 - 2x^2 - 2x = 0
x^3 - 3x^2 - 2x = 0
Factoring out an x:
x(x^2 - 3x - 2) = 0
Solving for x^2 - 3x - 2 = 0, we can use the quadratic formula:
x = (3 ± sqrt(3^2 - 4(1)(-2))) / 2
x = (3 ± sqrt(9 + 8)) / 2
x = (3 ± sqrt(17)) / 2
So, we have two x-values: (3 + sqrt(17)) / 2 and (3 - sqrt(17)) / 2.
To calculate the area, we need to find the integral of the upper curve (f(x)) and subtract the integral of the lower curve (g(x)):
Area = ∫[a, b] (f(x) - g(x)) dx
where a and b are the x-values of intersection.
The integral of f(x) = x^3 + x^2 can be found by using the power rule:
∫(x^3 + x^2) dx = (1/4)x^4 + (1/3)x^3
The integral of g(x) = 2x^2 + 2x can also be found by using the power rule:
∫(2x^2 + 2x) dx = (2/3)x^3 + x^2
Now, we can calculate the area:
Area = ∫[(3 - sqrt(17)) / 2, (3 + sqrt(17)) / 2] ((1/4)x^4 + (1/3)x^3 - (2/3)x^3 - x^2) dx
= ∫[(3 - sqrt(17)) / 2, (3 + sqrt(17)) / 2] ((1/4)x^4 - (1/3)x^3 - x^2) dx
Evaluating the definite integral:
Area = [(1/4)(1/5)x^5 - (1/3)(1/4)x^4 - (1/3)x^3] [(3 + sqrt(17)) / 2, (3 - sqrt(17)) / 2]
= [(1/20)x^5 - (1/12)x^4 - (1/3)x^3] [(3 + sqrt(17)) / 2, (3 - sqrt(17)) / 2]
Now you can plug in the x-values into the formula above and simplify to find the area. Please double-check your calculations and make sure you have substituted the correct values.