Find the arc length function for the curve y = 2x3/2 with starting point Po(25, 250) s(x) For the function f(x) = -e* + e *, prove that the arc length on any interval has the same value as the area under the curve. f(x) f'(x) 1[f'(x)]2 = 1 + = [f(x)]2 The arc length of the curve y = f(x) on the interval [a, b] is V1+f(x)2 dkVi d | f(x) dx, f(x) on the interval [a, b] which is the area under the curve y
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Solve parts these 2 parts of the question. Image attached.
![For the function f(x) = -e* + e *, prove that the arc length on any interval has the same value as the area under the curve.
f(x)
f'(x)
1[f'(x)]2 = 1 +
=
[f(x)]2
The arc length of the curve y = f(x) on the interval [a, b] is
V1+f(x)2 dkVi d |
f(x) dx,
f(x) on the interval [a, b]
which is the area under the curve y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5495962c-f5a5-4855-948a-0c159253a976%2F5f7d4f65-d300-46d0-96f8-969fb875e7b5%2Fexrerp.png&w=3840&q=75)
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