What is the arc length of $f(x)=xlnx$ in the interval $[1,e^2]$ ?

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Answer 1 · 2016-12-30T14:47:38

$16.168917 ...$

The Arc Length of curve $y=f(x)$ is calculated using the formula:

$L = int_a^b sqrt(1+f'(x)^2) \ dx$

So with $f(x)=xln(x)$ , we can apply the product rule to get:

$\ \ \ \ \ f'(x) = (x)(1/x) + (1)(lnx)$
$:. f'(x) = 1 + lnx$

And so the required Arc Length is given by:

$L = int_1^(e^2) sqrt(1+(1 + lnx)^2) \ dx$
$\ \ = int_1^(e^2) sqrt(1+(1 + 2lnx + (lnx)^2)) \ dx$
$\ \ = int_1^(e^2) sqrt(2 + 2lnx + (lnx)^2) \ dx$

Using Wolfram Alpha this integral evaluates to:

$L = int_1^(e^2) sqrt(2 + 2lnx + (lnx)^2) \ dx = 16.168917 ...$

What is the arc length of f(x)=xlnx f(x)=xlnx in the interval [1,e^2][1,e^2]?