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.
Between y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 for x in [0, 3]
find their intersection
x^2 - 4x + 1 = -x^2 + 4x - 5
2x^2 - 8x + 6 = 0
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
x = 1 or x = 3
look at
http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+%E2%88%92+4x+%2B+1+,+y+%3D+%E2%88%92x%5E2+%2B+4x+%E2%88%92+5
As you can see, they cross at (1,-2) and (3,-2)
but you want the area from 0 to 3.
you will have to do it two parts, from 0 to 1, and then from 1 to 3
I will do the second part, you do the first part, then add up the two areas.
from 1 to 3:
effective height = (-x^2+4x-5) - (x^2-4x+1)
= -2x^2 + 8x - 6
area = ∫(-2x^2 + 8x - 6) dx from 1 to 3
= [(-2/3)x^3 + 4x^2 - 6x] from 1 to 3
= ( (-2/3)(27) + 4(9) - 18) - ( (-2/3) + 4 - 6)
= 8/3
confirmation:
http://www.wolframalpha.com/input/?i=area+between+y+%3D+x%5E2+%E2%88%92+4x+%2B+1+,+y+%3D+%E2%88%92x%5E2+%2B+4x+%E2%88%92+5
now do the area from 0 to 1, and add to 8/3
notice the first graph is now above the second graph, so your height is
+2x^2 - 8x + 6
To find the area of the indicated region between the two curves, we will need to calculate the definite integral of the difference between the upper curve and the lower curve over the given interval.
First, let's graph the two curves to visualize the area we're trying to find. We can use graphing software or an online graphing tool for this.
The graph of the two curves, y = x^2 − 4x + 1 and y = −x^2 + 4x − 5, looks like this:
Graph:
Curve 1: y = x^2 − 4x + 1
Curve 2: y = −x^2 + 4x − 5
Next, we need to determine the points of intersection of the two curves. To find these points, we set the two equations equal to each other and solve for x:
x^2 − 4x + 1 = −x^2 + 4x − 5
Simplifying the equation, we get:
2x^2 - 8x + 6 = 0
Factoring the quadratic equation, we find:
(x - 1)(2x - 6) = 0
Setting each factor equal to zero:
x - 1 = 0, x = 1
2x - 6 = 0, x = 3
So the two curves intersect at x = 1 and x = 3.
Now, let's find the area between the curves using the definite integral:
A = ∫[0, 3] (upper curve - lower curve) dx
The upper curve is the parabola y = x^2 − 4x + 1, and the lower curve is the parabola y = −x^2 + 4x − 5.
A = ∫[0, 3] [(x^2 − 4x + 1) - (−x^2 + 4x − 5)] dx
= ∫[0, 3] [2x^2 - 8x + 6] dx
To evaluate this integral, use any integral calculator, math software, or a graphing calculator that can perform definite integrals.
Using technology to calculate the definite integral, we find that:
A ≈ 16.33 (rounded to two decimal places)
Therefore, the approximate area of the region between the curves y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 for x in [0, 3] is 16.33 square units.