By reversing the order of integration, evaluate the integral  
∫
6
0
∫
2
y
3
y
cos
(
x
3
−
6
)
d
x
d
y
 .

We are given the double integral:

\[
I = \int_0^6 \int_y^{2y} \cos(x^3 - 6) \, dx \, dy
\]

To evaluate this integral by reversing the order of integration, we need to first describe the limits of integration in terms of the region of integration.

### Step 1: Identify the Region of Integration
From the given limits:

- The outer integral has \( y \) going from 0 to 6: \( 0 \leq y \leq 6 \).
- The inner integral has \( x \) going from \( y \) to \( 2y \): \( y \leq x \leq 2y \).

Thus, the region \( R \) in the \( xy \)-plane is bounded by these inequalities:
\[
0 \leq y \leq 6 \quad \text{and} \quad y \leq x \leq 2y
\]

To reverse the order of integration, we need to express this region in terms of \( x \) first.

### Step 2: Rewrite the Region in Terms of \( x \)
For the region \( y \leq x \leq 2y \), we can describe the region as:

- \( x \) ranges from 0 to 12 (because when \( y = 6 \), \( 2y = 12 \)).
- For a given \( x \), \( y \) must satisfy \( \frac{x}{2} \leq y \leq x \).

Thus, the region in terms of \( x \) is:

\[
0 \leq x \leq 12 \quad \text{and} \quad \frac{x}{2} \leq y \leq x
\]

### Step 3: Set up the Reversed Integral
Now that we have the new limits of integration, we can rewrite the double integral as:

\[
I = \int_0^{12} \int_{\frac{x}{2}}^x \cos(x^3 - 6) \, dy \, dx
\]

### Step 4: Evaluate the Inner Integral
The inner integral with respect to \( y \) is straightforward since the integrand \( \cos(x^3 - 6) \) is independent of \( y \):

\[
\int_{\frac{x}{2}}^x \cos(x^3 - 6) \, dy = \cos(x^3 - 6) \left( x - \frac{x}{2} \right) = \cos(x^3 - 6) \frac{x}{2}
\]

### Step 5: Evaluate the Outer Integral
Substitute this into the outer integral:

\[
I = \int_0^{12} \cos(x^3 - 6) \frac{x}{2} \, dx
\]

Now, we have to evaluate the integral:

\[
I = \frac{1}{2} \int_0^{12} x \cos(x^3 - 6) \, dx
\]

### Step 6: Substitution to Simplify
Let \( u = x^3 - 6 \), so that \( du = 3x^2 \, dx \). To match the integrand, we need an \( x \, dx \), so rearranging:

\[
du = 3x^2 \, dx \quad \Rightarrow \quad \frac{du}{3} = x^2 \, dx
\]

This substitution would normally work, but it requires more detailed handling that could lead to numerical methods or further simplifications.

Would you like more detailed steps on this substitution, or should I continue with numeric methods?

---

### Questions:
1. What is the significance of changing the order of integration?
2. How does the substitution \( u = x^3 - 6 \) simplify the integral?
3. Can we estimate the result of the integral without solving it explicitly?
4. What types of functions are typically difficult to integrate directly?
5. How does the method of integration by parts compare to substitution for integrals like these?

**Tip**: Always sketch the region of integration when reversing the order to ensure you have the correct limits.

By reversing the order of integration, evaluate the integral ∫6 0 ∫2y y cos(x^3 − 6) dx dy.

Learn how to reverse the order of integration for the double integral ∫6 0 ∫2y y cos(x^3 − 6) dx dy, including step-by-step guidance on setting up the new limits, using substitution, and simplifying the integral. This problem involves concepts such as double integrals and trigonometric functions, suitable for college-level calculus.

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