Question

The value of the integral \( \iint_R (x+y) \,dy\,dx \) in the region R bounded by \( x=0, x=2, y=x, y=x+2 \), is

• A. 3
• B. 8
• C. 12
• D. 16

Hint:

For integrals over parallelograms not aligned with the axes, a change of variables can sometimes simplify the limits. For example, let \(u=x\) and \(v=y-x\). The limits would become \(0 \le u \le 2\) and \(0 \le v \le 2\). However, for simple polynomial integrands, direct integration as shown is often just as fast.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem asks to evaluate a double integral over a region R defined by four lines. The region is a parallelogram. We need to set up the iterated integral with the correct limits and then evaluate it.

Step 2: Key Formula or Approach:
The region R is defined by the inequalities \( 0 \le x \le 2 \) and \( x \le y \le x+2 \). The integral is set up as: \[ I = \int_{0}^{2} \int_{x}^{x+2} (x+y) \,dy \,dx \]
Step 3: Detailed Explanation:
First, we evaluate the inner integral with respect to \(y\), treating \(x\) as a constant: \[ \int_{x}^{x+2} (x+y) \,dy = \left[ xy + \frac{y^2}{2} \right]_{y=x}^{y=x+2} \] \[ = \left( x(x+2) + \frac{(x+2)^2}{2} \right) - \left( x(x) + \frac{x^2}{2} \right) \] \[ = \left( x^2 + 2x + \frac{x^2+4x+4}{2} \right) - \left( x^2 + \frac{x^2}{2} \right) \] \[ = \left( x^2 + 2x + \frac{x^2}{2} + 2x + 2 \right) - \frac{3x^2}{2} \] \[ = \left( \frac{3x^2}{2} + 4x + 2 \right) - \frac{3x^2}{2} = 4x + 2 \] Now, we evaluate the outer integral with respect to \(x\): \[ I = \int_{0}^{2} (4x+2) \,dx \] \[ = \left[ 4\frac{x^2}{2} + 2x \right]_{0}^{2} = \left[ 2x^2 + 2x \right]_{0}^{2} \] \[ = (2(2^2) + 2(2)) - (0) = (2(4) + 4) = 8 + 4 = 12 \]
Step 4: Final Answer:
The value of the integral is 12.