Questions: Let f be a continuous, positive, decreasing function on [3, ∞). Compare the values of the integral A = ∫ from 3 to 18 of f(t) dt and the series B = Σ from n=3 to 17 of f(n), C = Σ from n=4 to 18 of f(n).

Let f be a continuous, positive, decreasing function on [3, ∞). Compare the values of the integral

A = ∫ from 3 to 18 of f(t) dt

and the series

B = Σ from n=3 to 17 of f(n), C = Σ from n=4 to 18 of f(n).

Transcript text: Let $f$ be a continuous, positive, decreasing function on $[3, \infty)$. Compare the values of the integral \[ A=\int_{3}^{18} f(t) d t \] and the series \[ B=\sum_{n=3}^{17} f(n), \quad C=\sum_{n=4}^{18} f(n) . \]

Solution

Solution Steps

To compare the values of the integral $ A $ and the series $ B $ and $ C $, we can use the properties of integrals and sums for decreasing functions. For a decreasing function $ f $, the integral from $ a $ to $ b $ can be approximated by the sum of the function values at integer points. Specifically, the integral will be slightly less than the sum of the function values at the left endpoints and slightly more than the sum of the function values at the right endpoints.

Solution Approach

Calculate the integral $ A $ using numerical integration.
Calculate the sums $ B $ and $ C $ by summing the function values at the specified points.
Compare the values of $ A $, $ B $, and $ C $ to determine the correct inequality.

Step 1: Calculate the Integral $ A $

We computed the integral $ A = \int_{3}^{18} f(t) \, dt $ and found that

\[ A \approx 1.7918. \]

Step 2: Calculate the Series $ B $ and $ C $

Next, we calculated the series $ B = \sum_{n=3}^{17} f(n) $ and $ C = \sum_{n=4}^{18} f(n) $. The results were:

\[ B \approx 1.9396, \] \[ C \approx 1.6618. \]

Step 3: Compare the Values

Now, we compare the values of $ A $, $ B $, and $ C $: