Evaluate the following integral \( \iint(4 x+y)^{3} d A, R(0,2),(0,3) \)

To evaluate the double integral \[ \iint_R (4x + y)^3 \, dA \] over the region \( R \) defined by \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 \), follow these steps: ### Step 1: Set Up the Integral The region \( R \) is a rectangle with \( x \) ranging from 0 to 2 and \( y \) ranging from 0 to 3. Thus, the double integral can be expressed as: \[ \int_{0}^{2} \int_{0}^{3} (4x + y)^3 \, dy \, dx \] ### Step 2: Integrate with Respect to \( y \) First, integrate \( (4x + y)^3 \) with respect to \( y \): Let \( u = 4x + y \), hence \( du = dy \). The limits for \( u \) when \( y = 0 \) and \( y = 3 \) are \( u = 4x \) and \( u = 4x + 3 \), respectively. \[ \int_{0}^{3} (4x + y)^3 \, dy = \int_{4x}^{4x + 3} u^3 \, du = \left[ \frac{u^4}{4} \right]_{4x}^{4x + 3} = \frac{(4x + 3)^4 - (4x)^4}{4} \] ### Step 3: Integrate with Respect to \( x \) Now, integrate the result with respect to \( x \): \[ \int_{0}^{2} \frac{(4x + 3)^4 - (4x)^4}{4} \, dx = \frac{1}{4} \left( \int_{0}^{2} (4x + 3)^4 \, dx - \int_{0}^{2} (4x)^4 \, dx \right) \] Compute each integral separately: 1. **First Integral:** \[ \int_{0}^{2} (4x + 3)^4 \, dx \] Let \( u = 4x + 3 \), \( du = 4 \, dx \), so \( dx = \frac{du}{4} \). The limits change to \( u = 3 \) when \( x = 0 \) and \( u = 11 \) when \( x = 2 \). \[ \int_{3}^{11} u^4 \cdot \frac{du}{4} = \frac{1}{4} \left[ \frac{u^5}{5} \right]_3^{11} = \frac{11^5 - 3^5}{20} \] 2. **Second Integral:** \[ \int_{0}^{2} (4x)^4 \, dx = 4^4 \int_{0}^{2} x^4 \, dx = 256 \left[ \frac{x^5}{5} \right]_0^{2} = 256 \cdot \frac{32}{5} = \frac{8192}{5} \] ### Step 4: Combine the Results \[ \frac{1}{4} \left( \frac{11^5 - 3^5}{20} - \frac{8192}{5} \right) = \frac{11^5 - 3^5 - 32768}{20} \] Calculate the numerical values: \[ 11^5 = 161051, \quad 3^5 = 243 \] \[ 161051 - 243 - 32768 = 128040 \] \[ \frac{128040}{20} = 6402 \] ### Final Answer \[ \boxed{6402} \]