Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(y)=y(2-y), g(y)=-y $$

Step-by-Step Solution

Answer

The area of the region bounded by the curves \(y(2-y)\) and \(-y\) from y=0 to 2, is found by evaluating the integral \( \int_{0}^{2} [(y(2-y))-(-y)] dy \).

Step 1: Identify the points of intersection

The points of intersection are obtained by setting the two functions equal to each other and solving for y. So set \(y(2-y) = -y\). This will yield two solutions, which are y=0 and y=2. These are the values of y that the region of interest is between.

Step 2: Sketch the region

Sketch the graphs of both functions \(f(y)=y(2-y)\) and \(g(y)=-y\) between y=0 and y=2. This will give a visual representation of the region whose area needs to be found. \(f(y)\) will give a parabolic curve opening downwards with a maximum value at y=1, while \(g(y)\) will give a straight line with negative slope.

Step 3: Compute the area using integration

The area of region between two curves from a to b is found using the definite integral \( \int_{a}^{b} [f(y)-g(y)] dy \). Here, \(f(y)\) is the upper curve and \(g(y)\) is the lower curve. Since \(f(y)>g(y)\), substitute \( y(2-y) \) for \(f(y)\) and \(-y\) for \(g(y)\) resulting in the integral \( \int_{0}^{2} [(y(2-y))-(-y)] dy \). Evaluate this integral to get the area of the region.