To find the length of the curve described by the equation:

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

we will use the formula for the length of a curve $y(x)$ between two points $x = a$ and $x = b$, which is given by:

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

Step 1: Find $\frac{dy}{dx}$

The function $y(x)$ is defined as:

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

This is a definite integral with an upper limit as a variable, so by the Fundamental Theorem of Calculus, the derivative of this integral with respect to $x$ is:

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

Step 2: Set up the length formula

Now that we have $\frac{dy}{dx} = \sqrt{x^3 - 1}$, we substitute this into the arc length formula:

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

Simplifying inside the square root:

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

Step 3: Integrate

Now we compute the integral:

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

Evaluating from 1 to 4:

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

Since $4^{5/2} = (2^2)^{5/2} = 2^5 = 32$ and $1^{5/2} = 1$, we get:

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

Thus, the length of the curve is:

$L = \frac{62}{5} \approx 12.4.$

Final Answer:

The length of the curve is $\frac{62}{5}$, or approximately 12.4 units.

Would you like more details on any part of this solution?

Here are some questions you could explore further:

1. How does the Fundamental Theorem of Calculus apply in this context?
2. What is the geometric meaning of arc length in a curve?
3. How would the process change if the limits of integration were different?
4. Can you generalize this method for other integrals that define curves?
5. What are common mistakes when calculating the length of curves?

Tip: Always simplify the expression inside the square root before integrating when calculating arc lengths!