Problem 13 - College Algebra and Calculus: An Applied Approach

Question

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=x^{2}-x^{3} \quad[-1,0] $$

Step-by-Step Solution

Answer

The approximate area under the curve \(x^{2}-x^{3}\) over the interval [-1, 0] using the Midpoint Rule with 4 partitions is 0.23046875.

Step 1: Partition The Interval

The first step is to divide the interval [-1, 0] into 4 equal sub-intervals. The width of each sub-interval or the base of each rectangle, \(h\), would be \((b-a)/n = (0-(-1))/4 = 0.25\). The intervals then are [-1, -0.75], [-0.75, -0.5], [-0.5, -0.25], and [-0.25, 0].

Step 2: Calculate The Midpoints And Function Values

The next step is to calculate the midpoints of each interval (i.e. \(x_{i}\)) and the function values at each midpoint \(f(x_{i})\). They are as follows: At \(x_{-0.875}\), \(f(x)\) = (-0.875)^2 - (-0.875)^3 = 0.615234375, At \(x_{-0.625}\), \(f(x)\) = (-0.625)^2 - (-0.625)^3 = 0.244140625, At \(x_{-0.375}\), \(f(x)\) = (-0.375)^2 - (-0.375)^3 = 0.056640625, At \(x_{-0.125}\), \(f(x)\) = (-0.125)^2 - (-0.125)^3 = 0.005859375.

Step 3: Calculate The Area Of The Regions

The third step is to evaluate the function at the midpoint of each sub-interval and then calculate the area of each sub-interval. Each area can be calculated as \(h \times f(x_{i})\). The sum of these areas will provide the approximate area under the curve. The areas are: For x=-0.875, area = 0.25 * 0.615234375 = 0.15380859375, For x=-0.625, area = 0.25 * 0.244140625 = 0.06103515625, For x=-0.375, area = 0.25 * 0.056640625 = 0.01416015625, For x=-0.125, area = 0.25 * 0.005859375 = 0.00146484375

Step 4: Sum The Areas

The final step is to sum up the areas of all the rectangles to get the approximate area under the curve: approximate area = 0.15380859375 + 0.06103515625 + 0.01416015625 + 0.00146484375 = 0.23046875