Find the area of the indicated region: Between y = x2 − 5x + 3 and y = −x2 + 5x − 5 for x in [0, 4].
I keep ending up with 16/3, but this is incorrect.
check for intersection
x^2 - 5x + 3 = -x^2 + 5x - 5
2x^2 - 10x + 8 = 0
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or x = 4 but we need the area for
0 ? x ? 4 , so we have two sections, because there is a cross-over at x = 1
Area = ?(x^2 ? 5x + 3 + x^2 - 5x + 5)dx from 0 to 1 + ?(-x^2 + 5x-5 - x^2 + 5x - 3) dx from 1 to 4
= ? (2x^2 - 10x + 8)dx from 0 to 1 + ?(-x^2 + 10x - 8)dx from 1 to 4
= [(2/3)x^3 - 5x^2 + 8x)]from 0 to 1 + [(-2/3)x^3 + 5x^2 - 8x] form 1 to 4
= (2/3) - 5 + 8 - 0 + ( -128/3 + 80 - 32 - (-2/3 + 5 - 8) )
= 38/3
http://www.wolframalpha.com/input/?i=area+between+y+%3D+x%5E2+%E2%88%92+5x+%2B+3+and+y+%3D+%E2%88%92x%5E2+%2B+5x+%E2%88%92+5+from+0+to+4
Wolfram!!!! What a wonderful webpage
Thank you so much, especially for that website!!
To find the area of the region between the two given curves, we need to calculate the definite integral of the difference between the two functions over the given interval.
Let's first find the x-values where the two functions intersect. Set the two equations equal to each other and solve for x:
x^2 - 5x + 3 = -x^2 + 5x - 5
Combine like terms:
2x^2 - 10x + 8 = 0
Divide both sides by 2:
x^2 - 5x + 4 = 0
Factor the quadratic equation:
(x - 4)(x - 1) = 0
x = 4 and x = 1 are the x-values where the two functions intersect.
Now we can set up the integral to find the area between the curves. We need to split the integral into two parts: from 0 to 1 and from 1 to 4. This is because the curve y = x^2 - 5x + 3 is on top of the curve y = -x^2 + 5x - 5 in the first interval, and the curve y = -x^2 + 5x - 5 is on top in the second interval.
First, let's calculate the integral from 0 to 1:
∫[0 to 1] [(x^2 - 5x + 3) - (-x^2 + 5x - 5)] dx
Simplify the integrand:
∫[0 to 1] [2x^2 - 10x + 8] dx
Integrating, we get:
(2/3)x^3 - 5x^2 + 8x | [0 to 1]
Plug in the limits of integration:
[(2/3)(1)^3 - 5(1)^2 + 8(1)] - [(2/3)(0)^3 - 5(0)^2 + 8(0)]
Simplify:
(2/3 - 5 + 8) - (0 - 0 + 0) = (2/3 - 5 + 8)
Now, let's calculate the integral from 1 to 4:
∫[1 to 4] [(-x^2 + 5x - 5) - (x^2 - 5x + 3)] dx
Simplify the integrand:
∫[1 to 4] [-2x^2 + 10x - 8] dx
Integrating, we get:
-(2/3)x^3 + 5x^2 - 8x | [1 to 4]
Plug in the limits of integration:
-[(2/3)(4)^3 - 5(4)^2 + 8(4)] - [-(2/3)(1)^3 + 5(1)^2 - 8(1)]
Simplify:
-[(2/3)(64) - 5(16) + 32] + [(2/3) - 5 + 8]
Now, let's calculate the final result:
Result = (2/3 - 5 + 8) - [(2/3)(64) - 5(16) + 32] + [(2/3) - 5 + 8]
After simplifying the expression, the final result should be the correct area of the region between the two curves over the interval [0, 4].
To find the area of the region between the two curves, you need to first identify the points of intersection of the curves within the given interval [0, 4].
Let's solve the equations y = x^2 - 5x + 3 and y = -x^2 + 5x - 5 simultaneously to find the points of intersection:
x^2 - 5x + 3 = -x^2 + 5x - 5
Rearrange the equation to get:
2x^2 - 10x + 8 = 0
Divide the equation by 2:
x^2 - 5x + 4 = 0
Factor the quadratic equation:
(x - 4)(x - 1) = 0
This gives two possible values for x: x = 4 and x = 1.
Now we need to find the corresponding y-values for these x-values in both equations.
For y = x^2 - 5x + 3, if x = 4, then:
y = (4)^2 - 5(4) + 3 = 16 - 20 + 3 = -1
If x = 1, then:
y = (1)^2 - 5(1) + 3 = 1 - 5 + 3 = -1
For y = -x^2 + 5x - 5, if x = 4, then:
y = -(4)^2 + 5(4) - 5 = -16 + 20 - 5 = -1
If x = 1, then:
y = -(1)^2 + 5(1) - 5 = -1 + 5 - 5 = -1
As you can see, the two curves intersect at the points (1, -1) and (4, -1).
To find the area between these curves, you need to integrate the difference between the top curve (y = x^2 - 5x + 3) and the bottom curve (y = -x^2 + 5x - 5) over the interval [1, 4].
The area (A) can be calculated using the following integral:
A = ∫[a,b] (Top curve - Bottom curve) dx
In this case, a = 1 and b = 4. So the integral becomes:
A = ∫[1,4] [(x^2 - 5x + 3) - (-x^2 + 5x - 5)] dx
Simplifying this gives:
A = ∫[1,4] (2x^2 - 8) dx
Now, let's find the antiderivative of (2x^2 - 8):
A = (2/3)x^3 - 8x | from 1 to 4
Evaluating the integral from 1 to 4:
A = [(2/3)(4)^3 - 8(4)] - [(2/3)(1)^3 - 8(1)]
A = (128/3 - 32) - (2/3 - 8)
A = (128/3 - 96/3) - (-6/3)
A = (32/3) - (-6/3)
A = (32/3) + (6/3)
A = 38/3
Therefore, the correct answer for the area of the region between the two curves is 38/3.