# What is the arc length of #f(x)=x^2-2x+35# on #x in [1,7]#?

The arc length is

Arc length is given by:

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

Reverse the substitution:

Insert the limits of integration:

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To find the arc length ( L ) of the function ( f(x) = x^2 - 2x + 35 ) on the interval ([1, 7]), you can use the arc length formula:

[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} , dx ]

First, find the derivative ( f'(x) ) of ( f(x) ):

[ f'(x) = \frac{d}{dx}(x^2 - 2x + 35) ] [ f'(x) = 2x - 2 ]

Now, compute ( [f'(x)]^2 ):

[ [f'(x)]^2 = (2x - 2)^2 ] [ [f'(x)]^2 = 4x^2 - 8x + 4 ]

Next, compute ( \sqrt{1 + [f'(x)]^2} ):

[ \sqrt{1 + [f'(x)]^2} = \sqrt{1 + 4x^2 - 8x + 4} ] [ \sqrt{1 + [f'(x)]^2} = \sqrt{4x^2 - 8x + 5} ]

Now, integrate ( \sqrt{4x^2 - 8x + 5} ) over the interval ([1, 7]):

[ L = \int_{1}^{7} \sqrt{4x^2 - 8x + 5} , dx ]

This integral can be challenging to solve directly. You might need to use trigonometric substitution or other advanced integration techniques to evaluate this integral.

Once you have evaluated the integral over the interval ([1, 7]), you will get the arc length ( L ) of the function ( f(x) = x^2 - 2x + 35 ) on the interval ([1, 7]).

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