Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer.
Enclosed by y = x^2 − 4x + 1 and y = −x^2 + 4x − 5
the curves intersect at x=1,3 so the area is
∫[1,3] (−x^2 + 4x − 5 )-(x^2 − 4x + 1) dx
= ∫[1,3] -2x^2 + 8x - 6 dx
= -2/3 x^3 + 4x^2 - 6x [1,3]
= 8/3
thank you so much
To find the area of the region enclosed by the two curves y = x^2 − 4x + 1 and y = −x^2 + 4x − 5, we'll set up an integral to calculate the area between the two curves.
Step 1: Find the x-values of the points where the curves intersect.
Set the two equations equal to each other: x^2 − 4x + 1 = −x^2 + 4x − 5
Combine like terms: 2x^2 - 8x + 6 = 0
Divide both sides by 2: x^2 - 4x + 3 = 0
Factor the quadratic equation: (x - 1)(x - 3) = 0
Set each factor equal to zero and solve for x:
x - 1 = 0 => x = 1
x - 3 = 0 => x = 3
The two curves intersect at x = 1 and x = 3.
Step 2: Determine which curve is above the other in the enclosed region.
Evaluate the y-values of each curve at x = 1 and x = 3:
For y = x^2 − 4x + 1:
y(1) = 1^2 − 4(1) + 1 = -2
y(3) = 3^2 − 4(3) + 1 = 2
So, the curve y = x^2 − 4x + 1 is below the other curve in the region between x = 1 and x = 3.
Step 3: Set up the integral for the area between the curves.
The area between the curves can be calculated using the integral:
A = ∫[from 1 to 3] (−x^2 + 4x − 5) - (x^2 − 4x + 1) dx
Step 4: Evaluate the integral to find the area.
Let's simplify the equation to integrate:
A = ∫[from 1 to 3] -x^2 + 4x - 5 - x^2 + 4x - 1 dx
A = ∫[from 1 to 3] -2x^2 + 8x - 6 dx
Now, integrate the simplified equation:
A = [-2/3 * x^3 + 4x^2 - 6x] [from 1 to 3]
A = [-2/3 * (3)^3 + 4(3)^2 - 6(3)] - [-2/3 * (1)^3 + 4(1)^2 - 6(1)]
A = [-18 + 36 - 18] - [-2/3 + 4 - 6]
A = [0] - [-2/3 -2]
A = [0] - [-2/3 - 6/3]
A = [0] - [-8/3]
A = 8/3
Therefore, the area of the region enclosed by the curves y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 is 8/3 square units.
To find the area of the region enclosed by the two curves y = x^2 − 4x + 1 and y = −x^2 + 4x − 5, we can proceed by first finding the points where the two curves intersect.
Setting the two equations equal to each other, we get:
x^2 - 4x + 1 = -x^2 + 4x - 5
Now, we can rearrange this equation to form a quadratic equation:
2x^2 - 8x + 6 = 0
To find the solutions to this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -8, and c = 6. Plugging these values into the formula, we get:
x = (-(-8) ± √((-8)^2 - 4(2)(6))) / (2(2))
= (8 ± √(64 - 48)) / 4
= (8 ± √16) / 4
Simplifying further, we have:
x = (8 ± 4) / 4
So, the two possible x-values where the curves intersect are:
x = 3 or x = 1
To find the y-values corresponding to these x-values, we can substitute them back into one of the original equations. Let's use y = x^2 − 4x + 1:
For x = 3, y = (3)^2 - 4(3) + 1 = 9 - 12 + 1 = -2
For x = 1, y = (1)^2 - 4(1) + 1 = 1 - 4 + 1 = -2
Therefore, at x = 3 and x = 1, the two curves intersect at y = -2.
To visualize the graph, you can plot these two equations in a graphing calculator or any software that supports graphing functions. Looking at the graph, you will see that the parabola y = x^2 − 4x + 1 is above the parabola y = −x^2 + 4x − 5 in the given interval.
Now, to find the area enclosed by the curves, we need to integrate the difference between the two curves with respect to x.
∫[(x^2 − 4x + 1) - (−x^2 + 4x − 5)] dx
Simplifying this expression, we have:
∫[2x^2 - 8] dx
Integrating, we get:
(2/3)x^3 - 8x + C
To evaluate the definite integral of this expression over the interval where the curves intersect (from x = 1 to x = 3), we subtract the values of the antiderivative at the upper and lower limits:
[(2/3)(3^3) - 8(3)] - [(2/3)(1^3) - 8(1)]
= [(2/3)(27) - 24] - [(2/3)(1) - 8]
= [18 - 24] - [2/3 - 8]
= [-6] - [-22/3]
= -6 + 22/3
= (3/3)(-6) + 22/3
= -18/3 + 22/3
= 4/3
So, the area enclosed by the two curves is 4/3 square units.
To verify this answer using technology, you can use graphing software or online graphing calculators to plot the two equations and calculate the area.