Asked by Raj

Find the area of the region enclosed between the curves y=(x-2)^4 and y=(x-2)^3.
Answered by Damon
from x = what to x = whatever?
if x gets big you have x^4-x^3
as x ---> infinity the difference goes to infinity
Answered by Raj
The answer is 1/20.How?
Answered by Damon
perhaps you just mean the enclosed area between x = 0 and x = 2 ?
f(x) = (x-2)^3(x-3) =x^4-8x^3+27x^2-44x+24
integrate that from 0 to 2
(1/5)x^5 - 2x^4 + 9 x^3 - 22 x^2 + 24 x
at 2 = 32/5 - 32 + 72 - 88 + 48
= answer because 0 at 0
Answered by Damon
whoops, they cross at x = 2 but not at x = 0
at x = 3
(x-3) = 0 at x = 3
so do
(1/5)x^5 - 2x^4 + 9 x^3 - 22 x^2 + 24 x
at x = 3
and find difference from at x = 2
Answered by Damon
(1/5)(243) - 2(81) +9(27) - 22(9) +24(3)
Answered by Raj
Why no (x-2)^4-(x-2)^3
Answered by Damon
You can if you want but I could see where it was zero (at 2 and at 3) by factoring
Answered by Raj
From where did 3 come?
Answered by Damon
from f(x) = (x-2)^3(x-3)
because
(x-2)^4 - (x-2)^3 = (x-2)^3 [ (x-2)-1] = (x-2)^3 (x-3)
Answered by Raj
Thanks
Answered by Damon
You are welcome, just in time about 3 am in morning here.
Answered by Raj
I appreciate.
Answered by Reiny
They intersect at x = 2 and x = 3, and in that interval, y = (x-2)^3 is above y = (x-2)^4
so you want ∫(x-2)^3 - (x-2)^4 dx from 2 to 3
= [ (1/4)(x-2)^4 - (1/5)(x-2)^5] from 2 to 3
= ( (1/4)(1^5) - (1/5)(1^4) - (0 - 0)
= 1/4 - 1/5
= 1/20

https://www.wolframalpha.com/input/?i=%E2%88%AB%28x-2%29%5E3+-+%28x-2%29%5E4+dx+from+2+to+3
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