5.3: FTC Changed History
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 \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 \begin{align*}x\end{align*}axis.

Before doing this activity, students should understand that if \begin{align*}a < b\end{align*} and \begin{align*}f(x) > 0\end{align*}, then:
 \begin{align*}\int \limits_{a}^{b}f(x) > 0\end{align*}
 \begin{align*}\int \limits_{a}^{b}f(x) < 0\end{align*}
 \begin{align*}\int \limits_{b}^{a}f(x) < 0\end{align*}
 \begin{align*}\int \limits_{b}^{a}f(x) > 0\end{align*}
 Before starting this activity, students should go to the HOME screen and select \begin{align*}F6\end{align*} :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables.
Associated Materials
 Student Worksheet: FTC Changed History http://www.ck12.org/flexr/chapter/9729, scroll down to the third activity.
Problem 1 – Constant integrand
Students explore the function \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 \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.
\begin{align*}x\end{align*}  \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. \begin{align*}\int \limits_{0}^{0} 1.5 \ dt = 0\end{align*}; There is zero area under the graph of \begin{align*}y = 1.5\end{align*} from \begin{align*}x = 0\end{align*} to \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 \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 – NonConstant Integrand
Students investigate the behavior of \begin{align*}f(x) = \int \limits_{0}^{x} \frac{t}{2} \ dt\end{align*}. Students should note that this function changes at a nonconstant rate and are asked to explain why this is so (from a geometric point of view).
\begin{align*}x\end{align*}  \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. \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
\begin{align*}x\end{align*}  \begin{align*}\int \limits_{0}^{x} \frac{t^2  13t + 22}{9} dt\end{align*} 

1  \begin{align*}\frac{95}{54} = 1.76\end{align*} 
2  \begin{align*}\frac{62}{27} = 2.29\end{align*} 
3  \begin{align*}\frac{11}{6} = 1.83\end{align*} 
4  \begin{align*}\frac{16}{27} = 0.59\end{align*} 
5  \begin{align*}\frac{65}{54} = 1.20\end{align*} 
6  \begin{align*}\frac{10}{3} = 3.33\end{align*} 
7  \begin{align*}\frac{301}{54} = 5.57\end{align*} 
8  \begin{align*}\frac{208}{27} = 7.70\end{align*} 
9  \begin{align*}\frac{19}{2} = 9.50\end{align*} 
10  \begin{align*}\frac{290}{27} = 10.74\end{align*} 
11  \begin{align*}\frac{605}{54} = 11.20\end{align*} 
12  \begin{align*}\frac{32}{3} = 10.67\end{align*} 
13  \begin{align*}\frac{481}{54} = 8.91\end{align*} 
14  \begin{align*}\frac{154}{27} = 5.70\end{align*} 
The fractions are approximated to the nearest hundredth.
11. After \begin{align*}x = 2\end{align*}, the integral value begins to decrease.
12. The values for \begin{align*}x\end{align*} in which the integral decreases are \begin{align*}2 < x < 11\end{align*}; the function is negative.
13. The values for which the integral is increasing are \begin{align*}x < 2, x > 11\end{align*}; the function is positive.
14. The table seems to indicate \begin{align*}x = 11\end{align*}. To find out for sure, use fMin on the integral. You have to restrict the domain to \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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