Question

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ 2 x-3 y=0, x+y=5, y=0 $$

Short answer

The area of the region enclosed by the given equations is \( \frac{21}{2} \) square units

Step-by-step solution

Rewrite the equations

Rewrite each equation in the form y = f(x):\[\begin{align*} 2x - 3y &= 0 \Rightarrow y = \frac{2}{3}x \\x + y &= 5 \Rightarrow y = 5 - x \\y &= 0 \end{align*}\]

Sketch the region

Sketch the bounded region. It is a triangle with vertices at the points where the lines intersect. Solve the pairs of equations to find these intersection points: (0,0), (5,0), and (3,2).

Determine the limits of integration

The limits of integration correspond to the bounded region. The x-limits are 0 and 3 as per the intersections. The y-limits depend on x and are from the line y = 0 to the line y = 5 - x.

Setup the double integral

Now set up the double integral that represents the area of the region: \[\int_{0}^{3}\int_{0}^{5-x} dy dx = \int_{0}^{3}(5 - x) dx\]

Solve the integral

Solve the integral: \[\int_{0}^{3} (5 - x) dx = [5x - \frac{1}{2} x^2]_{0}^{3} = 15 - \frac{9}{2} = \frac{21}{2}\]