Problem 24 - 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. $$ f(y)=2 y, \quad[0,2] $$

Step-by-Step Solution

Answer

The approximate area using the midpoint rule with 4 sub-intervals is 2.5, while the exact area under the given function curve over the interval [0,2] is 4. The approximation therefore has a discrepancy compared to the exact value.

Step 1: Define the Function and Interval

The function given is \(f(y) = 2y\), and the interval \([a,b]\) over which we want to approximate the integral is \([0,2]\). The number of sub-intervals, \(n\), is given as 4.

Step 2: Divide Interval and Calculate Midpoints

Divide the interval \([0,2]\) into 4 equal parts. Each sub-interval will have a width of \(\Delta y = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5\). The midpoints of these intervals are then \(0.25, 0.75, 1.25, 1.75\).

Step 3: Apply the Midpoint Rule

Use the midpoint rule formula to approximate the integral, i.e., \(M_n = \Delta y [f(y_1) + f(y_2) + ... + f(y_n)]\), where \(y_i\) are the midpoints. This gives \(M_4 = 0.5[2(0.25) + 2(0.75) + 2(1.25) + 2(1.75)] = 2.5\)

Step 4: Find the Exact Integral

The exact area under the curve over the interval [0,2] is given by the definite integral: \(\int_0^2 2y dy = [y^2]_0^2 = 4 - 0 = 4\).

Step 5: Compare the Approximate Area with the Exact Area

Lastly, the approximate area using the midpoint rule is 2.5, while the exact area from integration is 4. Therefore, there is a discrepancy in this case.