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

Question

Isabella Ackerman · Accepted Answer

# 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 ... #

Emma Aldridge · Answer

undefined To find the arc length of $ f(x) = x \ln x $ on the interval $[1, e^2]$, you need to integrate the square root of $1 + (f'(x))^2$ over the given interval, where $f'(x)$ is the derivative of $f(x)$.

First, find the derivative of $ f(x) $:
$$ f'(x) = 1 + \ln x $$

Then, calculate the square root of $1 + (f'(x))^2$:
$$ \sqrt{1 + (1 + \ln x)^2} $$

Now, integrate this expression with respect to $x$ over the interval $[1, e^2]$:

$$ \int_{1}^{e^2} \sqrt{1 + (1 + \ln x)^2} \, dx $$

This integral represents the arc length of $ f(x) = x \ln x $ on the interval $[1, e^2]$. You can solve it using numerical methods or integration techniques.

Ezekiel Alexander · Answer

undefined To find the arc length of the curve $ f(x) = x \ln(x) $ on the interval $[1, e^2]$, you can use the formula for arc length:

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

First, find the derivative of $ f(x) $ which is $ f'(x) $. Then substitute $ f'(x) $ into the formula and integrate it over the given interval. This will give you the arc length of the curve on the specified interval.