What is the arclength of $f (x) = x^{2} e^{\frac{1}{x}}$ $f(x)=x^2e^(1/x)$ on $x \in [0, 1]$ $x in [0,1]$ ?

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What is the arclength of $f (x) = x^{2} e^{\frac{1}{x}}$ $f(x)=x^2e^(1/x)$ on $x \in [0, 1]$ $x in [0,1]$ ?

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Answer 1 · 2016-09-18T05:57:12

$\infty$ $oo$

Explanation:

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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What is the arclength of $f (x) = x^{2} e^{\frac{1}{x}}$ $f(x)=x^2e^(1/x)$ on $x \in [0, 1]$ $x in [0,1]$ ?

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What is the arclength of f(x)=x^2e^(1/x)f(x)=x2e1xf(x)=x^2e^(1/x) on x in [0,1]x∈[0,1]x in [0,1]?

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What is the arclength of $f (x) = x^{2} e^{\frac{1}{x}}$ $f(x)=x^2e^(1/x)$ on $x \in [0, 1]$ $x in [0,1]$ ?