To find the length of the curve given by the parametric equation

$y = \int_1^x \sqrt{t^3 - 1} \, dt, \quad 1 \leq x \leq 4,$

we will use the formula for the arc length of a curve defined parametrically:

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

Step-by-step Solution:

1. Compute $\frac{dy}{dx}$: Since $y$ is defined as an integral function, we can use the Fundamental Theorem of Calculus to differentiate it. This gives:

   $\frac{dy}{dx} = \sqrt{x^3 - 1}.$

2. Arc length formula: Now, substitute $\frac{dy}{dx}$ into the arc length formula:

   $L = \int_1^4 \sqrt{1 + \left( \sqrt{x^3 - 1} \right)^2} \, dx.$

3. Simplify the integrand: Simplifying the expression inside the square root:

   $\sqrt{1 + \left( \sqrt{x^3 - 1} \right)^2} = \sqrt{1 + (x^3 - 1)} = \sqrt{x^3}.$

   Thus, the length of the curve becomes:

   $L = \int_1^4 \sqrt{x^3} \, dx = \int_1^4 x^{3/2} \, dx.$

4. Integrate: Now, perform the integration:

   $\int x^{3/2} \, dx = \frac{2}{5} x^{5/2}.$

   Evaluate the definite integral from $1$ to $4$:

   $L = \left[ \frac{2}{5} x^{5/2} \right]_1^4 = \frac{2}{5} \left( 4^{5/2} - 1^{5/2} \right).$

   Calculate the values:

   $4^{5/2} = (2^2)^{5/2} = 2^5 = 32,$ and $1^{5/2} = 1$.

   Thus, the length of the curve is:

   $L = \frac{2}{5} \left( 32 - 1 \right) = \frac{2}{5} \times 31 = \frac{62}{5} = 12.4.$

So, the length of the curve is $\boxed{12.4}$.

Would you like more details or have any further questions?

Related Questions:

1. What is the general formula for arc length in parametric form?
2. How does the Fundamental Theorem of Calculus apply to the given problem?
3. Can the arc length formula be used for curves defined in polar coordinates?
4. How would the arc length formula change if $y$ was a function of $x$ directly rather than through an integral?
5. Could you solve the problem using a different method, like numerical integration?

Tip:

Remember that differentiating an integral function, as we did here, is a powerful tool provided by the Fundamental Theorem of Calculus.