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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 ddx0xf(t)dt=f(x)\begin{align*}\frac{d}{dx} \int \limits_{0}^{x}f(t)dt = f(x)\end{align*}.

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\begin{align*}x-\end{align*}axis.
• Before doing this activity, students should understand that if a<b\begin{align*}a < b\end{align*} and f(x)>0\begin{align*}f(x) > 0\end{align*}, then:
• abf(x)>0\begin{align*}\int \limits_{a}^{b}f(x) > 0\end{align*}
• abf(x)<0\begin{align*}\int \limits_{a}^{b}-f(x) < 0\end{align*}
• baf(x)<0\begin{align*}\int \limits_{b}^{a}f(x) < 0\end{align*}
• baf(x)>0\begin{align*}\int \limits_{b}^{a}-f(x) > 0\end{align*}
• Before starting this activity, students should go to the HOME screen and select F6\begin{align*}F6\end{align*} :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables.

Associated Materials

## Problem 1 – Constant integrand

Students explore the function 0x1.5 dt\begin{align*}\int \limits_{0}^{x} 1.5 \ dt\end{align*}. They should notice that there is a constant rate of change in the graph of f(x)=0x1.5 dt\begin{align*}f(x) = \int \limits_{0}^{x} 1.5 \ dt\end{align*} This rate of change is 1.5.

1. The table looks like the one below.

x\begin{align*}x\end{align*} 0x1.5 dt\begin{align*}\int \limits_{0}^{x}1.5 \ dt\end{align*}
1 1.5
2 3
3 4.5
4 6
5 7.5

2. 001.5 dt=0\begin{align*}\int \limits_{0}^{0} 1.5 \ dt = 0\end{align*}; There is zero area under the graph of y=1.5\begin{align*}y = 1.5\end{align*} from x=0\begin{align*}x = 0\end{align*} to x=0\begin{align*}x = 0\end{align*}.

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 x,0x0.5 dt\begin{align*}\left(x,\int \limits_{0}^{x} 0.5 \ dt \right)\end{align*}.

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)=0xt2 dt\begin{align*}f(x) = \int \limits_{0}^{x} \frac{t}{2} \ dt\end{align*}. 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\begin{align*}x\end{align*} 0xt2 dt\begin{align*}\int \limits_{0}^{x} \frac{t}{2} \ dt\end{align*}
1 0.25
2 1
3 2.25
4 4
5 6.25

8. 00t2 dt=0\begin{align*}\int \limits_{0}^{0} \frac{t}{2} \ dt = 0\end{align*} 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\begin{align*}x\end{align*} 0xt213t+229dt\begin{align*}\int \limits_{0}^{x} \frac{t^2 - 13t + 22}{9} dt\end{align*}
1 9554=1.76\begin{align*}\frac{95}{54} = 1.76\end{align*}
2 6227=2.29\begin{align*}\frac{62}{27} = 2.29\end{align*}
3 116=1.83\begin{align*}\frac{11}{6} = 1.83\end{align*}
4 1627=0.59\begin{align*}\frac{16}{27} = 0.59\end{align*}
5 6554=1.20\begin{align*}-\frac{65}{54} = -1.20\end{align*}
6 103=3.33\begin{align*}-\frac{10}{3} = -3.33\end{align*}
7 30154=5.57\begin{align*}-\frac{301}{54} = -5.57\end{align*}
8 20827=7.70\begin{align*}-\frac{208}{27} = -7.70\end{align*}
9 192=9.50\begin{align*}-\frac{19}{2} = -9.50\end{align*}
10 29027=10.74\begin{align*}-\frac{290}{27} = -10.74\end{align*}
11 60554=11.20\begin{align*}-\frac{605}{54} = -11.20\end{align*}
12 323=10.67\begin{align*}-\frac{32}{3} = -10.67\end{align*}
13 48154=8.91\begin{align*}-\frac{481}{54} = -8.91\end{align*}
14 15427=5.70\begin{align*}-\frac{154}{27} = -5.70\end{align*}

The fractions are approximated to the nearest hundredth.

11. After x=2\begin{align*}x = 2\end{align*}, the integral value begins to decrease.

12. The values for x\begin{align*}x\end{align*} in which the integral decreases are 2<x<11\begin{align*}2 < x < 11\end{align*}; the function is negative.

13. The values for which the integral is increasing are x<2,x>11\begin{align*}x < 2, x > 11\end{align*}; the function is positive.

14. The table seems to indicate x=11\begin{align*}x = 11\end{align*}. To find out for sure, use fMin on the integral. You have to restrict the domain to x>0\begin{align*}x > 0\end{align*}, 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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