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Handout Daily Prep 4.3 - The Definite Integral

Section Overview

In this section we introduce a notation used for expressing the limit of a Riemann sum. We thus define what a definite integral is. We will be discussing the definition, meaning, and use of the definite integral for the remainder of the course. Surprisingly, it has a very strong and natural connection to the derivative – a connection we will discuss and explore in Section 4.4. However, in this section, we deduce the properties that this new mathematical object, the definite integral, has based on its definition and its geometrical interpretation.

Section Basic learning objectives

These are the tasks you should be able to perform with reasonable fluency when you arrive at our next class meeting. Important new vocabulary words are indicated in italics.
  • Recognize the parts of the limit definition of the definite integral, especially how the definite integral results from taking a limit of Riemann sums.
  • Identify an integral sign, integrand, and limits of integration.
  • Explain what it means to evaluate a definite integral.
  • Interpret, geometrically, the quantity denoted by \(\displaystyle \int_{a}^{b} f(x) \ dx\text{.}\)
  • Recognize and apply properties that the definite integral possesses. Use these properties to evaluate definite integrals in special circumstances.

Section To prepare for class

Complete all actions listed below. Respond to the questions highlighted with Submit.

Checkpoint 120. πŸ“ [Submit] Evaluate Using Integral Properties.

Section After class

Solidifying the concepts discussed in class through practice is necessary to build your skills.

Section Advanced learning objectives

In addition to mastering the basic objectives, here are the tasks you should be able to perform after class, with practice:
  • Compute the average value of a function on the interval \([a,b]\) using \(\displaystyle \int_{a}^{b} f(x) \ dx\text{.}\)
  • Interpret the average value of a function as the height of a rectangle.

Section Additional suggestions

Section Answers

Subsection To prepare for class

    1. \(\displaystyle \displaystyle \frac{5}{4}\)
    2. \(\displaystyle \displaystyle \frac{\pi+1}{2}\)
  1. \(L_{4} = (20)(3) + (10)(3) + (5)(3) + (2)(3) = 111, R_{4} = 54\text{;}\) these average to 82.5.
    1. \(\displaystyle \displaystyle a=1, b=4, n=3, \Delta x_{k} = 1, x_{k}^{*} = k+\frac{3}{4}\)
    2. \(\displaystyle \displaystyle a=0.5, b=3.5, n=6, \Delta x_{k} = 0.5, x_{k}^{*} = \frac{1}{2}k\)

Subsection Additional suggestions

  1. This is a right sum \(R_{3}\text{.}\)
    Here, \(\displaystyle a=0.5, b=5, n=3, \Delta x_{k} = \frac{3}{2}\) and
    \(\displaystyle x_{k}^{*} = \frac{1}{2}+ k\left(\frac{3}{2}\right)\text{.}\) The sum is
    \(\displaystyle \frac{4}{1+2}(1.5) + \frac{4}{1+3.5}(1.5) + \frac{4}{1+5}(1.5) = 4.\overline{3}\text{.}\)
  2. Here \(\displaystyle \Delta x_{k} = \frac{6}{4}\text{,}\) \(\displaystyle x_{k}^{*} = -\frac{1}{4}+ k\left(\frac{6}{4}\right)\text{.}\)
    So, \(\displaystyle \sum_{k=1}^{3}\frac{4}{1+\left[ -\frac{1}{4}+k\left(\frac{6}{4}\right) \right]}\cdot \left(\frac{6}{4}\right)\text{.}\)

Subsection And yet additional suggestions

  1. \(\displaystyle \int_{0}^{90}C(t) \ dt\) represents the total cost to heat house for the first 90 days of 2020 (in dollars).
    \(\displaystyle \frac{1}{90-0}\int_{0}^{90}C(t) \ dt\) represents the average cost (dollars per day) of heating the house for the first 90 days of 2020.
  2. \(\displaystyle \frac{1}{40-0}\int_{0}^{40}V(t) \ dt = \frac{1}{40}\int_{0}^{40}225(1.15)^{t} \ dt \approx \$10,740.46\text{.}\) The Average Value of a Function applet produces a nice way to see this as illustrated in FigureΒ 124.
    Graph of an exponential function f(x) that increases very rapidly on the interval from about x = 20 to x = 40. The curve starts near the x‑axis on the left and rises steeply to a large value by x = 40. The region under the curve from x = 20 to x = 40 is shaded in blue to represent the integral, with the label β€œintegral β‰ˆ 129618.37.” A horizontal red line labeled β€œaverage value: y = 10740.46” crosses the graph. A text box near the curve displays the formula f(x) = 225Β·1.15^x. Grid lines and axes with tick marks are shown.
    Figure 124. The average value of \(y=225(1.15)^x\) on \([0,40]\text{.}\)