Problem 26 - College Algebra and Calculus: An Applied Approach

Question

Use the Midpoint Rule with \(n=4\) to approximate the area of the region. Compare your result with the exact area obtained with a definite integral. $$ f(y)=4 y-y^{2}, \quad[0,4] $$

Step-by-Step Solution

Answer

The approximate area calculated using the Midpoint Rule is 10 while the actual area calculated using definite integral is 10.67.

Step 1: Set Up the Midpoint Rule Formula

The Midpoint Rule is given by the formula \(\Delta y \sum_{i=1}^{n} f(M_i)\) where \(\Delta y = \frac{\text{upper-bound - lower-bound}}{n}\), \(M_i\) are midpoints of each subinterval and the function \(f(x)\) is evaluated at these midpoints. For this problem, we have \(\Delta y=\frac{4-0}{4}=1\). So, the midpoints \(M_i\) are \(0.5, 1.5, 2.5, 3.5\).

Step 2: Perform the Midpoint Rule Calculation

Plug \(M_i\) values into the function, then sum them up and multiply by \(\Delta y\). That is, sum \(f(0.5), f(1.5), f(2.5)\), and \(f(3.5)\), then multiply the result by \(\Delta y (=1)\). After performing the calculations, the Midpoint approximation results in an area of 10.

Step 3: Calculate the Definite Integral

The definite integral over the interval [0,4] can be computed as follows: \(\int_{0}^{4} (4y - y^2) dy\). Solving this yields an actual area of 10.67.

Step 4: Compare the Results

The Midpoint approximation calculated an area of 10 while the definite integral calculated an area of 10.67. This shows that while the Midpoint Rule provides a good approximate value, it slightly underestimates the true value of the integral.