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# 5.3: FTC Changed History

Difficulty Level: At Grade 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.

Feb 23, 2012

Nov 04, 2014