Find the exact length of the curvey = x-x2 +sin-1(V 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) = e+ e~X f'(x) = |1[f'(x)]2 = 1=[f(x)]2The arc length of the curve y = f(x) on the interval [a, b] isfx)12 dx =L =f(x) dx,1 [f'(x)]2 dx =Vwhich is the area under the curve y f(x) on the interval [a, b]

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Asked Oct 13, 2019
Find the exact length of the curve
y = x-x2 +sin-1(V
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Find the exact length of the curve y = x-x2 +sin-1(V

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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) = e+ e~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
fx)12 dx =
L =
f(x) dx,
1 [f'(x)]2 dx =
V
which is the area under the curve y f(x) on the interval [a, b]
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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) = e+ e~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 fx)12 dx = L = f(x) dx, 1 [f'(x)]2 dx = V which is the area under the curve y f(x) on the interval [a, b]

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check_circleExpert Solution
Step 1

Given:

y= vx-x + sin"
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y= vx-x + sin"

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Step 2

Formula used:

dy
Arc length
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dy Arc length

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Step 3

Now to start, at first differentiating the given func...

d
x-x'+sin"(Vx)
dx
dy d
sin (x)
Vx-x+
dx
dy
d
x
1
1
x-x dc
dy
1
(1-2x)
1
1
(x)
2Vx) dx
dy
1
1
1
-2x
2x
dy
1-2x
1
2Vx-x2
2/x-x
dx
dy
1-
1
dx
2x-x
2Vx-x
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d x-x'+sin"(Vx) dx dy d sin (x) Vx-x+ dx dy d x 1 1 x-x dc dy 1 (1-2x) 1 1 (x) 2Vx) dx dy 1 1 1 -2x 2x dy 1-2x 1 2Vx-x2 2/x-x dx dy 1- 1 dx 2x-x 2Vx-x

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Math

Calculus

Derivative

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