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What is the arclength of #f(x)=x^2e^(1/x)# on #x in [0,1]#?

Answer 1

#oo#

Without calculation, we can see easily that #lim_(x->0^+)x^2e^(1/x)=oo#, meaning #f(x)# has a vertical asymptote at #x=0#, and so the arclength must be infinite on #[0, 1]#.

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Answer 2

Evelyn Agard

To find the arc length of the function ( f(x) = x^2e^{1/x} ) on the interval ( [0, 1] ), you would need to use the formula for arc length:

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

Where ( f'(x) ) is the derivative of ( f(x) ) with respect to ( x ).

First, find the derivative ( f'(x) ), then plug it into the formula along with the limits of integration ( a = 0 ) and ( b = 1 ), and integrate.

The derivative ( f'(x) ) can be found using the product rule and chain rule:

[ f'(x) = 2xe^{1/x} - \frac{1}{x^3} e^{1/x} ]

Now, plug ( f'(x) ) into the arc length formula:

[ L = \int_{0}^{1} \sqrt{1 + \left(2xe^{1/x} - \frac{1}{x^3} e^{1/x}\right)^2} , dx ]

Integrate this expression within the given limits to find the arc length of ( f(x) ) on the interval ( [0, 1] ).

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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.