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
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.
2 answers
Thank you so much, especially for that website!!