Question

The value of \(\int_{0}^{4} \int_{0}^{4-x} x y \, dx \, dy\) is:

• A. 544/3
• B. 32/3
• C. 17/3
• D. 16/3

Hint:

Exam Tip:
For double integrals over triangular regions:

• Identify the order of integration correctly.
• The limits of the inner integral may depend on the outer variable.
• Be careful with algebra.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate a double integral over a triangular region. The limits of integration are: \(0 \le x \le 4\), \(0 \le y \le 4 - x\).

Step 2: Key Formula or Approach:
\[ \int_{0}^{4} \int_{0}^{4-x} x y \, dy \, dx \]

Step 3: Detailed Explanation:
First, integrate with respect to \(y\): \[ \int_0^{4-x} y \, dy = \left[ \frac{y^2}{2} \right]_0^{4-x} = \frac{(4-x)^2}{2} \] Now, integrate with respect to \(x\): \[ \int_0^4 x \cdot \frac{(4-x)^2}{2} \, dx = \frac{1}{2} \int_0^4 x(16 - 8x + x^2) \, dx = \frac{1}{2} \int_0^4 (16x - 8x^2 + x^3) \, dx \] \[ = \frac{1}{2} \left[ 8x^2 - \frac{8}{3}x^3 + \frac{1}{4}x^4 \right]_0^4 \] \[ = \frac{1}{2} \left[ 8(16) - \frac{8}{3}(64) + \frac{1}{4}(256) \right] = \frac{1}{2} \left[ 128 - \frac{512}{3} + 64 \right] = \frac{1}{2} \left[ 192 - \frac{512}{3} \right] = \frac{1}{2} \left[ \frac{576 - 512}{3} \right] = \frac{1}{2} \cdot \frac{64}{3} = \frac{32}{3} \]

Step 4: Final Answer:
Therefore, option (B) is correct.