Question

peer instruction: inclass_nov21
time to make your 2nd vote!
suppose ( f(x) ) is decreasing on ( a, b ). if we use a left - endpoint riemann sum to
approximate the value of ( int_{a}^{b} f(x) d x ), the result will be:
choose one
a. exactly right
b. an underestimate
c. an overestimate
d. wrong on so many levels

Explanation:

Step1: Recall the concept of left - endpoint Riemann sum

For a function \(y = f(x)\) on the interval \([a,b]\) divided into \(n\) sub - intervals \([x_{i},x_{i + 1}]\) with \(\Delta x=\frac{b - a}{n}\) and \(x_{i}=a + i\Delta x\), the left - endpoint Riemann sum is \(L_{n}=\sum_{i = 0}^{n-1}f(x_{i})\Delta x\).

Step2: Analyze the function \(y = f(x)\) which is decreasing on \([a,b]\)

Since \(f(x)\) is decreasing on \([a,b]\), for the sub - interval \([x_{i},x_{i+1}]\), \(f(x_{i})\geq f(x)\) for all \(x\in[x_{i},x_{i + 1}]\). Then \(\int_{x_{i}}^{x_{i+1}}f(x)dx\leq f(x_{i})\Delta x\) (because the area under the curve \(y = f(x)\) on \([x_{i},x_{i+1}]\) is less than or equal to the area of the rectangle with height \(f(x_{i})\) and width \(\Delta x\)).

Step3: Sum up for the entire interval \([a,b]\)

\(\int_{a}^{b}f(x)dx=\sum_{i = 0}^{n - 1}\int_{x_{i}}^{x_{i+1}}f(x)dx\leq\sum_{i=0}^{n - 1}f(x_{i})\Delta x\)

Answer:

C. An overestimate