Question

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ g(x)=\frac{4}{2-x}, \quad y=4, \quad x=0 $$

Short answer

The bounded area is 4 square units.

Step-by-step solution

Sketch the functions

It's important to understand what the functions look like on a graph. So, draw the functions \(g(x)=\frac{4}{2-x}\), \(y=4\), and the vertical line \(x=0\) (y-axis). Here, \(y=4\) is a horizontal line meanwhile \(g(x)=\frac{4}{2-x}\) is a hyperbola that opens to the left since the denominator subtracts x.

Find intersection points

Now, find the points where the graphs intersect each other as these will be the boundaries of the area. Set the equations to each other and solve: \(\frac{4}{2-x}=4\). After solving this equation, one gets \(x=1\) as the intersection point.

Setup the integral for the area

To calculate the area, you will subtract one function from another and integrate between the intersection points. The area A is given by the integral \[A=\int_{a}^{b}(f(x)-g(x))\, dx\] Integrate from x=0 (y-axis) to x=1 (intersection point). Here, \(f(x)=4\) (top curve) and \(g(x)=\frac{4}{2-x}\) (bottom curve). Thus, the integral becomes \[A=\int_{0}^{1}(4-\frac{4}{2-x})\, dx\]

Evaluate the integral

Next, perform the integration. The integral of a constant is the constant times x and for the second term one needs to apply the natural logarithm rule in integral calculus. \[A=4x-4\ln |2-x|\] Evaluating this between the limits 0 to 1 gives the bounded area as 4 units square.