#Calculus Integration Help Needed
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tulip marsh
Well, what's given is just a function y = f(x) such that f'(x) = √(x^2 - 4x + 3), f((3) = 0.
Recall that the length element of y = f(x) is:
dl = √(dx^2 + dy^2) = √(1 + (dy/dx)^2)dx = √(1 + f'(x)^2)dx
As we've shown above, f'(x) = √(x^2 - 4x + 3). So:
dl = √(1 + x^2 - 4x + 3)dx = √(x^2 - 4x + 4)dx = √((x - 2)^2)dx = |x - 2|dx
So, to find the length, integrate dl from 4 to 8. Note that |x - 2| = x - 2 for x > 2.
south quiver
Well, what's given is just a function y = f(x) such that f'(x) = √(x^2 - 4x + 3)...
So it doesn't matter that the question has 3 as the lower limit, when using the arclength formula the upper limit we use is 8 and the lower is 4?
tulip marsh
So it doesn't matter that the question has 3 as the lower limit, when using the ...
Yes.
Basically, whether the lower limit is 3 or anything else, this just moves f(x) up or down, which obviously doesn't change its length in the given limits.
Well, to be precise, √((x - 2)^2) = |x - 2|. But since our region is all to the right of x = 2, it will be equal to (x - 2) there.
Other than that, everything's good. Though, I would've taken the antiderivative as (1/2)(x - 2)^2. But that doesn't matter.
south quiver
Ok cool, thanks a lot. You are very helpful. Appreciate it!
tulip marsh
Thanks 😄
You're welcome!
south quiver
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