To find the arc length of the curve $y = x - \ln{x}$ on the interval $[1, 4]$, we use the formula for the arc length of a curve:

$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step 1: Find the derivative of $y$

Given $y = x - \ln{x}$, we first find $\frac{dy}{dx}$:

$\frac{dy}{dx} = 1 - \frac{1}{x}$

Step 2: Square the derivative

Now, square the derivative:

$\left( \frac{dy}{dx} \right)^2 = \left( 1 - \frac{1}{x} \right)^2 = 1 - 2\frac{1}{x} + \frac{1}{x^2}$

Step 3: Plug into the arc length formula

Substitute this into the arc length formula:

$L = \int_1^4 \sqrt{1 + 1 - 2\frac{1}{x} + \frac{1}{x^2}} \, dx$

Simplify the expression under the square root:

$L = \int_1^4 \sqrt{2 - 2\frac{1}{x} + \frac{1}{x^2}} \, dx$

Step 4: Simplify further

Factor out a 2:

$L = \int_1^4 \sqrt{\frac{(x - 1)^2}{x^2}} \, dx$

This simplifies to:

$L = \int_1^4 \frac{|x - 1|}{x} \, dx$

Since $x \geq 1$ in the interval $[1, 4]$, we can drop the absolute value:

$L = \int_1^4 \frac{x - 1}{x} \, dx = \int_1^4 \left( 1 - \frac{1}{x} \right) \, dx$

Step 5: Compute the integral

Now, integrate each term:

$L = \int_1^4 1 \, dx - \int_1^4 \frac{1}{x} \, dx$

[ L = \left[ x \right]_1^4 - \left[ \ln{x} \right]_1^4 ]

$L = (4 - 1) - (\ln{4} - \ln{1})$

$L = 3 - \ln{4}$

Since $\ln{1} = 0$, the final arc length is:

$L = 3 - \ln{4}$

Thus, the arc length of the curve is approximately $L \approx 3 - 1.386 = 1.614$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you find the derivative of logarithmic functions?
2. Can the arc length formula be applied to parametric curves?
3. How would the result change if the interval was different, say [1, 5]?
4. What happens if the curve's equation changes to $y = x + \ln{x}$?
5. Can arc length be calculated using numerical methods for complex curves?

Tip: When simplifying expressions inside a square root, look for opportunities to factor or simplify terms before integrating to avoid complex computations.