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

To evaluate the given integral \( \iint_R (4x + y)^3 \, dA \) over the region \( R = [0, 2] \times [0, 3] \), we can set it up as a double integral: \[ \int_{0}^{2} \int_{0}^{3} (4x + y)^3 \, dy \, dx. \] First, we integrate with respect to \( y \): \[ \int_{0}^{3} (4x + y)^3 \, dy. \] We can use substitution to simplify this integral. Let \( u = 4x + y \), so \( du = dy \). When \( y = 0 \), \( u = 4x \), and when \( y = 3 \), \( u = 4x + 3 \). Adjusting the integral, we get: \[ \int_{4x}^{4x + 3} u^3 \, du. \] Now, we evaluate \( \int u^3 \, du \): \[ \frac{u^4}{4} \Bigg|_{4x}^{4x + 3} = \frac{(4x + 3)^4}{4} - \frac{(4x)^4}{4}. \] Calculating \( (4x + 3)^4 - (4x)^4 \): Using the binomial expansion: \[ (4x + 3)^4 = \sum_{k=0}^{4} \binom{4}{k} (4x)^{4-k}(3)^k = (4x)^4 + 4(4x)^3(3) + 6(4x)^2(3^2) + 4(4x)(3^3) + (3^4). \] Now we can simplify \( (4x + 3)^4 - (4x)^4 \), gathering the terms: \[ 4 \cdot 4^3 x^3 \cdot 3 + 6 \cdot 16 x^2 \cdot 9 + 4 \cdot 4 x \cdot 27 + 81 = 192x^3 + 864x^2 + 108x + 81. \] Therefore, we have: \[ \int_{0}^{3} (4x + y)^3 \, dy = \frac{1}{4} \left( (192 x^3 + 864 x^2 + 108x + 81) \right). \] Next, we integrate that result with respect to \( x \): \[ \int_{0}^{2} (48x^3 + 216x^2 + 27x + \frac{81}{4}) \, dx. \] Calculating this integral piece by piece: - \( \int 48x^3 \, dx = 12x^4 \Big|_0^2 = 12(16) = 192 \) - \( \int 216x^2 \, dx = 72x^3 \Big|_0^2 = 72(8) = 576 \) - \( \int 27x \, dx = \frac{27}{2}x^2 \Big|_0^2 = 2 \cdot 27 = 54 \) - \( \int \frac{81}{4} \, dx = \frac{81}{4}x \Big|_0^2 = 2 \cdot \frac{81}{4} = \frac{81}{2} \) Putting it all together: \[ 192 + 576 + 54 + \frac{81}{2} = 822 + \frac{81}{2} = 822 + 40.5 = 862.5. \] Thus, the value of the integral is \[ \boxed{862.5}. \]