Math Problem Statement

Find \int_{0}^{1} \int_{2y}^{2} cosx^2 dxdy by changing the order of integration

Solution

To solve the double integral 012y2cos(x2)dxdy\int_{0}^{1} \int_{2y}^{2} \cos(x^2) \, dx \, dy by changing the order of integration, we'll first need to understand the region of integration and then change the order accordingly.

Step 1: Describe the Region of Integration

The integral is given as:

012y2cos(x2)dxdy\int_{0}^{1} \int_{2y}^{2} \cos(x^2) \, dx \, dy

The inner integral limits 2y2y to 2 indicate the bounds for xx, while the outer integral limits 0 to 1 indicate the bounds for yy.

For yy from 0 to 1:

  • xx ranges from 2y2y to 2.

This describes a region in the xyxy-plane where:

  1. 0y10 \leq y \leq 1,
  2. 2yx22y \leq x \leq 2.

Step 2: Expressing yy in Terms of xx

Next, let's express yy in terms of xx:

  • From x=2yx = 2y, we get y=x2y = \frac{x}{2}.
  • yy ranges from 0 to 1, so xx ranges from 0 to 2.

For fixed xx between 0 and 2:

  • yy ranges from 00 to x2\frac{x}{2}.

Step 3: Write the Integral with the New Order

Changing the order of integration, the integral becomes:

020x2cos(x2)dydx\int_{0}^{2} \int_{0}^{\frac{x}{2}} \cos(x^2) \, dy \, dx

Step 4: Simplify and Evaluate

Notice that cos(x2)\cos(x^2) does not depend on yy, so the inner integral is straightforward:

02[cos(x2)y]y=0y=x2dx\int_{0}^{2} \left[ \cos(x^2) \cdot y \right]_{y=0}^{y=\frac{x}{2}} \, dx

This simplifies to:

02x2cos(x2)dx\int_{0}^{2} \frac{x}{2} \cos(x^2) \, dx

Finally, the integral is:

1202xcos(x2)dx\frac{1}{2} \int_{0}^{2} x \cos(x^2) \, dx

Step 5: Perform the Substitution

To solve this integral, use the substitution u=x2u = x^2, hence du=2xdxdu = 2x \, dx.

1202xcos(x2)dx=1404cos(u)du\frac{1}{2} \int_{0}^{2} x \cos(x^2) \, dx = \frac{1}{4} \int_{0}^{4} \cos(u) \, du

Evaluating this:

14[sin(u)]04=14(sin(4)sin(0))=sin(4)4\frac{1}{4} \left[ \sin(u) \right]_{0}^{4} = \frac{1}{4} (\sin(4) - \sin(0)) = \frac{\sin(4)}{4}

So, the final result is:

sin(4)4\frac{\sin(4)}{4}

Final Answer:

012y2cos(x2)dxdy=sin(4)4\int_{0}^{1} \int_{2y}^{2} \cos(x^2) \, dx \, dy = \frac{\sin(4)}{4}

Would you like more details or have any questions?

Related Questions:

  1. How do you perform a change of variables in double integrals?
  2. What are the common methods for evaluating double integrals?
  3. How does the region of integration affect the order of integration?
  4. Can you explain the geometric interpretation of changing the order of integration?
  5. What are some common trigonometric integrals and their solutions?

Tip: When changing the order of integration, always sketch the region to clearly understand the limits.

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Math Problem Analysis

Mathematical Concepts

Calculus
Double Integrals
Trigonometric Integration

Formulas

\(\int \cos(x^2) \, dx\)
Substitution method: \(u = x^2\)

Theorems

Fubini's Theorem

Suitable Grade Level

College Level (Calculus II or III)