Question

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

Short answer

The value of the double integral over the described region is -25.5

Step-by-step solution

Understanding the Limits of Integration

The given double integral is in the form of \(\int_{a}^{b} \int_{g(y)}^{h(y)} f(x, y) dx dy\), where a = 0, b = 6 are the definite limits for y, and \(g(y) = y/2\), \(h(y) = 3\) are the limits for x. y varies between 0 and 6, and x varies between y/2 and 3.

Sketching the Region of Integration

Plot the line \(x = y/2\), which is a line inclined at an angle of 45 degree and passing through origin and the line \(x = 3\) which is a vertical line. The area enclosed between these two lines and y=0 and y=6 is the region of integration. For y=0 and y=6, the value of x would be 0 and 3 respectively. Hence, the region of integration is a strip in the first quadrant of the xy-plane.

Evaluate the Inner Integral

Evaluate the inner integral first, which is with respect to x. Find an antiderivative of the expression \(x + y\), which is \(0.5x^2 + xy\). At each x, we substitute the upper and lower limits of x, respectively, which gives \( [0.5*3^2 + 3y] - [0.5*(y/2)^2 + y^2/2] = 3.75 + 2.5y - 0.25y^2 \).

Evaluate the Outer Integral

Now we evaluate the remaining integral, integrating with respect to y. Find the antiderivative of the expression \(3.75 + 2.5y - 0.25y^2\), which is \(3.75y + 1.25y^2 - 0.083y^3\). Then substituting the limits of y, which gives \( [3.75*6 + 1.25*6^2 - 0.083*6^3] - [3.75*0 + 1.25*0^2 - 0.083*0^3] = 37.5 + 45 - 108 = -25.5 \)