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Let's break down how to determine which integral represents the length of the curve given by \(x = y + y^3\) over the interval \(1 \leq y \leq 4\).

### Curve Length Formula
The arc length \(L\) of a curve given by \(x = f(y)\) from \(y = a\) to \(y = b\) is:

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

Here, \(x = y + y^3\). So we need to compute \(\frac{dx}{dy}\) and substitute it into the integral.

### Step 1: Compute \(\frac{dx}{dy}\)
\[
\frac{dx}{dy} = \frac{d}{dy} \left( y + y^3 \right) = 1 + 3y^2
\]

### Step 2: Plug \(\frac{dx}{dy} = 1 + 3y^2\) into the Length Formula
\[
L = \int_1^4 \sqrt{1 + \left( 1 + 3y^2 \right)^2} \, dy
\]

### Step 3: Identify the Correct Integral from the Options
We need to find an option that matches the integral:

\[
\int_1^4 \sqrt{1 + \left( 1 + 3y^2 \right)^2} \, dy
\]

Looking at the given options, **Option E** matches this expression exactly:

\[
\boxed{E: \int_1^4 \sqrt{1 + \left( 1 + 3y^2 \right)^2} \, dy}
\]

### Final Answer
The correct choice is:

\[
\boxed{E}
\]

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Do you want further clarifications or a detailed walk-through of another step?

### Five Related Questions:
1. How do we find arc length for parametric curves?
2. What if the function was given as \(y = f(x)\)? How would the arc length formula change?
3. What are some practical applications of arc length in physics and engineering?
4. How would the arc length formula change if the curve were given in polar coordinates?
5. How does numerical integration apply when finding arc lengths?

### Tip:
When calculating arc length, it’s essential to simplify the derivative expression as much as possible before substituting it into the integral to make integration easier.

Which of the following integrals represents the length of the curve x = y + y^3, 1 ≤ y ≤ 4?

This problem asks to find the length of the curve x = y + y^3 for the interval 1 ≤ y ≤ 4 using calculus. It includes calculating the derivative of the curve and using the arc length formula. Suitable for students in Grades 11-12.

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