\( \int \int ( 4 x + y ) ^ { 3 } d A , R ( 0,2 ) , ( 0,3 ) \)

To evaluate the double integral \[ \iint_R (4x + y)^3 \, dA, \] where the region \( R \) is defined by \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 \), follow these steps: ### Step 1: Set Up the Integral The integral can be written as: \[ \int_{x=0}^{2} \int_{y=0}^{3} (4x + y)^3 \, dy \, dx \] ### Step 2: Evaluate the Inner Integral (with respect to \( y \)) \[ \int_{0}^{3} (4x + y)^3 \, dy \] Let \( u = 4x + y \). Then, \( du = dy \), and when \( y = 0 \), \( u = 4x \); when \( y = 3 \), \( u = 4x + 3 \). \[ \int_{4x}^{4x+3} u^3 \, du = \frac{u^4}{4} \Big|_{4x}^{4x+3} = \frac{(4x + 3)^4 - (4x)^4}{4} \] ### Step 3: Substitute Back into the Outer Integral \[ \int_{0}^{2} \frac{(4x + 3)^4 - (4x)^4}{4} \, dx = \frac{1}{4} \left( \int_{0}^{2} (4x + 3)^4 \, dx - \int_{0}^{2} (4x)^4 \, dx \right) \] ### Step 4: Evaluate Each Integral Separately 1. **First Integral:** Let \( u = 4x + 3 \), then \( du = 4 \, dx \), so \( dx = \frac{du}{4} \). When \( x = 0 \), \( u = 3 \); when \( x = 2 \), \( u = 11 \). \[ \int_{0}^{2} (4x + 3)^4 \, dx = \frac{1}{4} \int_{3}^{11} u^4 \, du = \frac{1}{20} \left[ u^5 \right]_{3}^{11} = \frac{11^5 - 3^5}{20} \] 2. **Second Integral:** \[ \int_{0}^{2} (4x)^4 \, dx = 4^4 \int_{0}^{2} x^4 \, dx = 256 \left[ \frac{x^5}{5} \right]_{0}^{2} = 256 \times \frac{32}{5} = \frac{8192}{5} \] ### Step 5: Combine the Results \[ \frac{1}{4} \left( \frac{11^5 - 3^5}{20} - \frac{8192}{5} \right) \] To simplify: \[ \frac{1}{4} \left( \frac{11^5 - 3^5 - 32768}{20} \right) = \frac{11^5 - 3^5 - 32768}{80} \] ### Step 6: Compute the Numerical Value Calculate \( 11^5 \), \( 3^5 \), and substitute: \[ 11^5 = 161051, \quad 3^5 = 243 \] \[ \frac{161051 - 243 - 32768}{80} = \frac{128040}{80} = 1600.5 \] ### Final Answer \[ \iint_R (4x + y)^3 \, dA = 1600.5 \] --- **Summary:** The value of the double integral is **1600.5**.