Question

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$

Short answer

The region of integration is a rectangle in the xy-plane with vertices at the points (0,0), (1,0), (0,3), and (1,3). The value of the double integral over this region is 18.

Step-by-step solution

Sketch the region of integration

The region of integration is defined by the limits of the double integral, \(0 \leq y \leq 1\) and \(0 \leq x \leq 3\). The graph of this region is a rectangle in the xy-plane with vertices at points (0,0), (1,0), (0,3), and (1,3).

Set up the integral

The integrand function is \(2x + 6y\). The order of integration is \( dy dx\). Therefore, the double integral is written as \[\int_{0}^{3} \int_{0}^{1}(2x + 6y) dy dx.\]

Evaluate the inner integral

The inner integral is \[\int_{0}^{1}(2x + 6y) dy.\]The variable of integration is y, while x is treated as a constant. Integrating gives \(2xy + 3y^2\). Then, evaluate this at y=0 and y=1, subtract the two results to obtain \(2x + 3 - 0 = 2x + 3\).

Evaluate the outer integral

The outer integral is \[\int_{0}^{3}(2x + 3) dx.\]This yields \(x^2 + 3x\), evaluated between 0 and 3. Subtracting the two results gives \(9 + 9 - 0 = 18\).