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5.3: FTC Changed History

Created by: CK-12

ID: 9778

Time required: 45 minutes

Activity Overview

This activity builds student comprehension of functions defined by a definite integral, where the independent variable is an upper limit of integration. Students are led to the brink of a discovery of a discovery of the Fundamental Theorem of Calculus, that \frac{d}{dx} \int \limits_{0}^{x}f(t)dt = f(x).

Topic: Fundamental Theorem of Calculus

  • Graph a function and use Measurement > Integral to estimate the area under the curve in a given interval.
  • Use Integral (in the Calculus menu) to obtain the exact value of a definite integral.

Teacher Preparation and Notes

  • This investigation should follow coverage of the definition of a definite integral, and the relationship between the integral of a function and the area of a region bounded by the graph of a function and the x-axis.
  • Before doing this activity, students should understand that if a < b and f(x) > 0, then:
    • \int \limits_{a}^{b}f(x) > 0
    • \int \limits_{a}^{b}-f(x) < 0
    • \int \limits_{b}^{a}f(x) < 0
    • \int \limits_{b}^{a}-f(x) > 0
  • Before starting this activity, students should go to the HOME screen and select F6 :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables.

Associated Materials

Problem 1 – Constant integrand

Students explore the function \int \limits_{0}^{x} 1.5 \ dt. They should notice that there is a constant rate of change in the graph of f(x) = \int \limits_{0}^{x} 1.5 \ dt This rate of change is 1.5.

1. The table looks like the one below.

x \int \limits_{0}^{x}1.5 \ dt
1 1.5
2 3
3 4.5
4 6
5 7.5

2. \int \limits_{0}^{0} 1.5 \ dt = 0; There is zero area under the graph of y = 1.5 from x = 0 to x = 0.

3. 1.5 units

4. The graph will be a line through the origin with slope 1.5.

Students will enter their data into the lists and then view the graph of \left(x,\int \limits_{0}^{x} 0.5 \ dt \right).

5. A line; yes (student answers may vary)

6. The same as before, except the slope would be 0.5 instead of 1.5.

Problem 2 – Non-Constant Integrand

Students investigate the behavior of f(x) = \int \limits_{0}^{x} \frac{t}{2} \ dt. Students should note that this function changes at a non-constant rate and are asked to explain why this is so (from a geometric point of view).

x \int \limits_{0}^{x} \frac{t}{2} \ dt
1 0.25
2 1
3 2.25
4 4
5 6.25

8. \int \limits_{0}^{0} \frac{t}{2} \ dt = 0 The height and the length of the triangle are 0 so the area is 0.

9. The area changes by a different amount each time because both the height and width are increasing.

10. The graph is not linear. It is a parabola as seen by the formula in the above screen shot.

Problem 3 – An Integrand That Changes Sign

x \int \limits_{0}^{x} \frac{t^2 - 13t + 22}{9} dt
1 \frac{95}{54} = 1.76
2 \frac{62}{27} = 2.29
3 \frac{11}{6} = 1.83
4 \frac{16}{27} = 0.59
5 -\frac{65}{54} = -1.20
6 -\frac{10}{3} = -3.33
7 -\frac{301}{54} = -5.57
8 -\frac{208}{27} = -7.70
9 -\frac{19}{2} = -9.50
10 -\frac{290}{27} = -10.74
11 -\frac{605}{54} = -11.20
12 -\frac{32}{3} = -10.67
13 -\frac{481}{54} = -8.91
14 -\frac{154}{27} = -5.70

The fractions are approximated to the nearest hundredth.

11. After x = 2, the integral value begins to decrease.

12. The values for x in which the integral decreases are 2 < x < 11; the function is negative.

13. The values for which the integral is increasing are x < 2, x > 11; the function is positive.

14. The table seems to indicate x = 11. To find out for sure, use fMin on the integral. You have to restrict the domain to x > 0, to get the answer.

15. Yes, we have seen a similar situation. The minimum occurs on a function where the function stops decreasing and starts increasing.

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Date Created:

Feb 23, 2012

Last Modified:

Feb 23, 2012
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TI.MAT.ENG.TE.1.Calculus.5.3

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