Calculus
posted by Diana .
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
(x1)(x3) = 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+4x5)  (x^24x+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
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