What is the arc length of #f(x)=x^2/12 + x^(-1)# on #x in [2,3]#?

Answer 1

Elijah Abram

#approx 1.03425#

#f'(x)=1/6x-1/x^2# so our integral will be

#int_2^3sqrt(1+(1/6*x-1/x^2)^2)dx approx 1.03425#

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Answer 2

Emily Ammerman

To find the arc length of the function ( f(x) = \frac{x^2}{12} + \frac{1}{x} ) on the interval ([2,3]), you would use the formula for arc length:

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

Where ( f'(x) ) is the derivative of ( f(x) ).

First, find ( f'(x) ): [ f'(x) = \frac{1}{6}x - \frac{1}{x^2} ]

Now, plug ( f'(x) ) into the formula: [ L = \int_{2}^{3} \sqrt{1 + \left(\frac{1}{6}x - \frac{1}{x^2}\right)^2} , dx ]

Now, integrate to find the arc length.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.