Math Problem Statement

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

Solution

We are given the double integral:

I=06y2ycos(x36)dxdyI = \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 yy going from 0 to 6: 0y60 \leq y \leq 6.
  • The inner integral has xx going from yy to 2y2y: yx2yy \leq x \leq 2y.

Thus, the region RR in the xyxy-plane is bounded by these inequalities: 0y6andyx2y0 \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 xx first.

Step 2: Rewrite the Region in Terms of xx

For the region yx2yy \leq x \leq 2y, we can describe the region as:

  • xx ranges from 0 to 12 (because when y=6y = 6, 2y=122y = 12).
  • For a given xx, yy must satisfy x2yx\frac{x}{2} \leq y \leq x.

Thus, the region in terms of xx is:

0x12andx2yx0 \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=012x2xcos(x36)dydxI = \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 yy is straightforward since the integrand cos(x36)\cos(x^3 - 6) is independent of yy:

x2xcos(x36)dy=cos(x36)(xx2)=cos(x36)x2\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=012cos(x36)x2dxI = \int_0^{12} \cos(x^3 - 6) \frac{x}{2} \, dx

Now, we have to evaluate the integral:

I=12012xcos(x36)dxI = \frac{1}{2} \int_0^{12} x \cos(x^3 - 6) \, dx

Step 6: Substitution to Simplify

Let u=x36u = x^3 - 6, so that du=3x2dxdu = 3x^2 \, dx. To match the integrand, we need an xdxx \, dx, so rearranging:

du=3x2dxdu3=x2dxdu = 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=x36u = 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.

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Order of Integration
Trigonometric Functions
Substitution Method

Formulas

Double integral formula: ∫∫ f(x,y) dx dy
Substitution: u = x^3 - 6

Theorems

Fubini's Theorem for reversing the order of integration

Suitable Grade Level

College-level Calculus