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The problem asks to evaluate the iterated integral:

\[
I = \int_0^1 \int_{1-y}^{1+y} \left( 6y^2 + 14x \right) dx \, dy
\]

### Step-by-Step Solution:

1. **Inner integral** (with respect to \( x \)):
   
   For the inner integral, we are integrating with respect to \( x \) first, while treating \( y \) as a constant. The inner integral is:

   \[
   \int_{1-y}^{1+y} \left( 6y^2 + 14x \right) dx
   \]

   Split this into two integrals:

   \[
   = \int_{1-y}^{1+y} 6y^2 \, dx + \int_{1-y}^{1+y} 14x \, dx
   \]

   - For the first integral \( \int_{1-y}^{1+y} 6y^2 \, dx \), since \( 6y^2 \) is a constant with respect to \( x \), the result is:

     \[
     6y^2 \cdot \left( (1+y) - (1-y) \right) = 6y^2 \cdot (2y) = 12y^3
     \]

   - For the second integral \( \int_{1-y}^{1+y} 14x \, dx \), apply the power rule:

     \[
     \left[ 7x^2 \right]_{1-y}^{1+y} = 7 \left( (1+y)^2 - (1-y)^2 \right)
     \]

     Use the difference of squares formula:

     \[
     (1+y)^2 - (1-y)^2 = \left( (1+y) - (1-y) \right) \cdot \left( (1+y) + (1-y) \right) = 2y \cdot 2 = 4y
     \]

     So the result of the second integral is:

     \[
     7 \cdot 4y = 28y
     \]

2. **Result of the inner integral:**

   The result of the inner integral is:

   \[
   12y^3 + 28y
   \]

3. **Outer integral** (with respect to \( y \)):

   Now integrate the result of the inner integral with respect to \( y \):

   \[
   \int_0^1 (12y^3 + 28y) \, dy
   \]

   Split this into two integrals:

   \[
   \int_0^1 12y^3 \, dy + \int_0^1 28y \, dy
   \]

   - For the first integral \( \int_0^1 12y^3 \, dy \), apply the power rule:

     \[
     \left[ 12 \cdot \frac{y^4}{4} \right]_0^1 = 3y^4 \Big|_0^1 = 3(1)^4 - 3(0)^4 = 3
     \]

   - For the second integral \( \int_0^1 28y \, dy \), apply the power rule:

     \[
     \left[ 28 \cdot \frac{y^2}{2} \right]_0^1 = 14y^2 \Big|_0^1 = 14(1)^2 - 14(0)^2 = 14
     \]

4. **Final result:**

   Adding the two results together:

   \[
   I = 3 + 14 = 17
   \]

So, the value of the iterated integral is \( \boxed{17} \).

---

Do you have any questions or would you like more details? Here are some related questions:

1. How does the order of integration affect the result in an iterated integral?
2. Can you explain how the limits of integration are determined for double integrals?
3. What is the difference between a definite and an indefinite integral in terms of application?
4. How does the power rule apply when integrating polynomials?
5. How would this problem change if the integrand involved trigonometric functions?

**Tip:** Always verify the limits of integration before starting to ensure the correct region is being integrated over.

Evaluate the iterated integral I = ∫[0 to 1] ∫[1-y to 1+y] (6y^2 + 14x) dx dy

Learn how to evaluate the iterated integral ∫[0 to 1] ∫[1-y to 1+y] (6y^2 + 14x) dx dy with a step-by-step breakdown of polynomial integration and iterated integrals. Suitable for undergraduate calculus students.

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