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

Question

Ezekiel Alexander · Accepted Answer

#approx 16.4316#

We are using the Formula

#int_a^bsqrt(1+(f'(x))^2)dx# Calculating #f'(x)#

#f'(x)=2xe^(1/x)+x^2*e^(1/x)(-1/x^2)# simplifying we get

#f'(x)=e^(1/x)(2x-1)# so we have to compute

#int_1^2sqrt(1+(e^(1/x)(2x-1))^2)dx#

This integral can not expressed by the known elementary functions.

By a numerical method we get #approx 16.4316#

Charles Arocha · Answer

undefined To find the arc length of the function $ f(x) = x^2e^{1/x} $ on the interval $[1,2]$, we use the formula for arc length, which is given by:

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

Where $ f'(x) $ represents the derivative of the function $ f(x) $.

First, we need to find the derivative of $ f(x) $, which involves using the product rule and the chain rule. After finding $ f'(x) $, we substitute it into the arc length formula.

The derivative of $ f(x) $ is given by:

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

Now, we substitute $ f'(x) $ into the arc length formula:

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

After substituting, we integrate the expression from $ x = 1 $ to $ x = 2 $ to find the arc length $ L $. This integral may be evaluated numerically using numerical integration techniques or appropriate software.

Once the integral is evaluated, the resulting value will be the arc length of the function $ f(x) = x^2e^{1/x} $ on the interval $[1,2]$.