1. The Fundamental Theorem of Calculus
Chapter 6.4

2. Fundamental Theorem, Part I
The two parts of the Fundamental Theorem of Calculus finally tie together the solutions to the tangent line problem and the area problem
The first part says that the definite integral of a continuous function is a differentiable function of its upper limit of integration, as well as what that derivative is
You saw this informally in Exploration 2 from the last section, where you were able to show that 𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡⟹ 𝑑𝐹 𝑑𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓 𝑡 𝑑𝑡=𝑓(𝑥)

3. Fundamental Theorem, Part I

4. Fundamental Theorem, Part I
The second part says that the definite integral of a continuous function from 𝑎 to 𝑏 can be found from any one of the function’s antiderivatives 𝐹 as the number 𝐹 𝑏 −𝐹(𝑎)
In the proof of the first part of the theorem that follows, keep in mind that
We will show that, if 𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡, then 𝑑𝐹 𝑑𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡=𝑓(𝑥)
The proof requires the MVT for integrals, which says that, at some value in an interval [𝑎,𝑏], a function takes on its average value on the interval

5. Theorem 4: the Fundamental Theorem of Calculus, Part I
If 𝑓 is continuous on [𝑎,𝑏], then the function
𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
has a derivative at every point 𝑥 in [𝑎,𝑏], and
𝑑𝐹 𝑑𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡=𝑓(𝑥)

6. Theorem 4: the Fundamental Theorem of Calculus, Part I
PROOF
We first apply the definition of the derivative to F
𝑑𝐹 𝑑𝑥 = lim ℎ→0 𝐹 𝑥+ℎ −𝐹 𝑥 ℎ
Now, since 𝐹 𝑥+ℎ = 𝑎 𝑥+ℎ 𝑓(𝑡) 𝑑𝑡 and 𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡, then
𝑑𝐹 𝑑𝑥 = lim ℎ→0 𝑎 𝑥+ℎ 𝑓 𝑡 𝑑𝑡− 𝑎 𝑥 𝑓 𝑡 𝑑𝑡 ℎ

7. Theorem 4: the Fundamental Theorem of Calculus, Part I
PROOF We can transform this to 𝑑𝐹 𝑑𝑥 = lim ℎ→0 𝑎 𝑥+ℎ 𝑓 𝑡 𝑑𝑡+ 𝑥 𝑎 𝑓 𝑡 𝑑𝑡 ℎ = lim ℎ→0 1 ℎ 𝑥 𝑥+ℎ 𝑓(𝑡) 𝑑𝑡 Since ℎ=𝑥+ℎ−𝑥, then this last expression is the average value of the function over the interval from 𝑥 to 𝑥+ℎ. By the MVT for integrals, then, there exists a value 𝑐 in [𝑥,𝑥+ℎ] such that 𝑓 𝑐 = 1 ℎ 𝑥 𝑥+ℎ 𝑓(𝑡) 𝑑𝑡

8. Theorem 4: the Fundamental Theorem of Calculus, Part I
PROOF Our equation now becomes 𝑑𝐹 𝑑𝑥 = lim ℎ→0 𝑓(𝑐) As demonstrated in the next slide, as ℎ→0, the value of 𝑓(𝑐) approaches 𝑓(𝑥) (because 𝑓 is continuous), so we conclude that 𝑑𝐹 𝑑𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡=𝑓(𝑥)

9. Theorem 4: the Fundamental Theorem of Calculus, Part I

10. Theorem 4: the Fundamental Theorem of Calculus, Part I
We can interpret this theorem as follows
It says that every continuous function 𝑓 is the derivative of some other function, 𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
It says that the processes of integration and differentiation are inverses of one another
Also remember that 𝐹 in the above equation is itself a function
It is important to keep this in mind because you will come across functions for which there is no known antiderivative
You will learn to use 𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡 as a function in its own right

11. Example 1: Applying the Fundamental Theorem
Find 𝑑 𝑑𝑥 −𝜋 𝑥 cos 𝑡 𝑑𝑡 and 𝑑 𝑑𝑥 0 𝑥 1 1+ 𝑡 2 𝑑𝑡

12. Example 1: Applying the Fundamental Theorem
Find
𝑑 𝑑𝑥 −𝜋 𝑥 cos 𝑡 𝑑𝑡 and 𝑑 𝑑𝑥 0 𝑥 𝑡 2 𝑑𝑡
𝑑 𝑑𝑥 −𝜋 𝑥 cos 𝑡 𝑑𝑡= cos 𝑥 and 𝑑 𝑑𝑥 0 𝑥 𝑡 2 𝑑𝑡= 1 1+ 𝑥 2

13. The Fundamental Theorem and the Chain Rule
For some integrals, the upper and lower limits of integration are themselves function
We apply the chain rule in these cases in the following way
Suppose that 𝑢(𝑥) is a function of 𝑥 and 𝐹(𝑢)= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡
Then by the chain rule, 𝑑𝐹 𝑑𝑥 = 𝑑𝐹 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑥

14. The Fundamental Theorem and the Chain Rule
Suppose that 𝑢(𝑥) is a function of 𝑥 and 𝐹(𝑢)= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡
Then by the chain rule, 𝑑𝐹 𝑑𝑥 = 𝑑𝐹 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑥
So we have
𝑑𝐹 𝑑𝑥 = 𝑑 𝑑𝑢 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡⋅ 𝑑𝑢 𝑑𝑥 =𝑓 𝑢 𝑥 ⋅ 𝑑𝑢 𝑑𝑥

15. The Fundamental Theorem and the Chain Rule
More generally, suppose the upper limit of integration is a function 𝑢(𝑥) and the lower limit is a function 𝑣(𝑥), as shown below
𝑦= 𝑣 𝑥 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡
We can choose any value 𝑎 in the domain of 𝑓 and use the Additivity Rule
𝑦= 𝑣 𝑥 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡+ 𝑣 𝑥 𝑎 𝑓(𝑡) 𝑑𝑡

16. The Fundamental Theorem and the Chain Rule
𝑦= 𝑣 𝑥 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡+ 𝑣 𝑥 𝑎 𝑓(𝑡) 𝑑𝑡
Now, rewrite the right-hand integral and use the Chain Rule twice
𝑦= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡+ 𝑣 𝑥 𝑎 𝑓(𝑡) 𝑑𝑡= 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡− 𝑎 𝑣 𝑥 𝑓(𝑡) 𝑑𝑡
𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑢 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡− 𝑑 𝑑𝑣 𝑎 𝑣 𝑥 𝑓(𝑡) 𝑑𝑡

17. The Fundamental Theorem and the Chain Rule
𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑢 𝑎 𝑢 𝑥 𝑓(𝑡) 𝑑𝑡− 𝑑 𝑑𝑣 𝑎 𝑣 𝑥 𝑓(𝑡) 𝑑𝑡
Applying the Chain Rule we finally get
𝑑𝑦 𝑑𝑥 =𝑓 𝑢 𝑥 ⋅ 𝑢 ′ 𝑥 −𝑓 𝑣 𝑥 ⋅𝑣′(𝑥)

18. Example 2: The Fundamental Theorem With the Chain Rule
Find the derivative of 𝑦= 1 𝑥 2 cos 𝑡 𝑑𝑡.

19. Example 2: The Fundamental Theorem With the Chain Rule
Find the derivative of 𝑦= 1 𝑥 2 cos 𝑡 𝑑𝑡.
Here we have that 𝑢 𝑥 = 𝑥 2 . We get
𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑢 1 𝑥 2 cos 𝑡 𝑑𝑡= cos 𝑥 2 ⋅2𝑥=2𝑥⋅ cos 𝑥 2

20. Example 3: Variable Lower Limits of Integration
Find the derivatives of
𝑦= 𝑥 5 3𝑡 sin 𝑡 𝑑𝑡
𝑦= 2𝑥 𝑥 𝑒 𝑡 𝑑𝑡

21. Example 3: Variable Lower Limits of Integration
𝑦= 𝑥 5 3𝑡 sin 𝑡 𝑑𝑡
Apply the Additivity Rule
𝑦=− 5 𝑥 3𝑡 sin 𝑡 𝑑𝑡
𝑑𝑦 𝑑𝑥 =− 𝑑 𝑑𝑥 5 𝑥 3𝑡 sin 𝑡 𝑑𝑡=−3𝑥 sin 𝑥

22. Example 3: Variable Lower Limits of Integration
𝑦= 2𝑥 𝑥 𝑒 𝑡 𝑑𝑡
Apply the Chain Rule twice
𝑦= 0 𝑥 𝑒 𝑡 𝑑𝑡− 0 2𝑥 𝑒 𝑡 𝑑𝑡
𝑑𝑦 𝑑𝑥 = 𝑑 𝑑𝑥 0 𝑥 𝑒 𝑡 𝑑𝑡− 𝑑 𝑑𝑥 0 2𝑥 𝑒 𝑡 𝑑𝑡= 2𝑥 2+ 𝑒 𝑥 2 − 2 2+ 𝑒 2𝑥

23. Example 4: Constructing a Function with a Given Derivative and Value
Find a function 𝑦=𝐹(𝑥) with derivative 𝑑𝑦 𝑑𝑥 = tan 𝑥 that satisfies the condition 𝐹 3 =5.

24. Example 4: Constructing a Function with a Given Derivative and Value
Although the textbook doesn’t yet use differentials, it is helpful to rewrite this first in differential form 𝑑𝑦= tan 𝑥 𝑑𝑥 This can serve to remind you that you must take the integral to find the function. 𝐹 𝑥 =∫𝑑𝑦= 3 𝑥 tan 𝑡 𝑑𝑡 or 𝐹 𝑥 =𝑦= 3 𝑥 tan 𝑡 𝑑𝑡

25. Example 4: Constructing a Function with a Given Derivative and Value
𝐹 𝑥 =𝑦= 3 𝑥 tan 𝑡 𝑑𝑡 Now consider that 𝐹 3 = 3 3 tan 𝑡 𝑑𝑡=0. So if we want 𝐹 3 =5, then our function is 𝐹 𝑥 =5+ 3 𝑥 tan 𝑡 𝑑𝑡 Note that this is an initial value problem. Always remember to write the initial x- value as the lower limit of integration and the value of the function added to the integral.

26. What is the Function?
In later chapters you will learn how to find the antiderivative of tan 𝑥 , which is
5+ 3 𝑥 tan 𝑡 𝑑𝑡=5+ ln cos 3 cos 𝑥
However, your calculator can evaluate this integral in its integral form, without having to find the antiderivative
It does this by approximation methods (like the RRAM and LRAM)
For example, use your calculator to find 𝐹 2 = tan 𝑡 𝑑𝑡
Your answer should be approximately

27. The Effect of Changing a in 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
By the FTC1 (Fundamental Theorem of Calculus Part I), the value of 𝑎 in 𝑑 𝑑𝑥 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡 does not affect the result
However, the value of 𝑎 does affect 𝑦=𝐹(𝑥) and its graph in 𝐹 𝑥 = 𝑎 𝑥 𝑓 𝑡 𝑑𝑡
To see how, what is the value of 𝐹(𝑎)?
𝐹 𝑎 = 𝑎 𝑎 𝑓(𝑡) 𝑑𝑡=0

28. The Effect of Changing a in 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
To see how, what is the value of 𝐹(𝑎)?
𝐹 𝑎 = 𝑎 𝑎 𝑓(𝑡) 𝑑𝑡=0
Hence, the graph of 𝐹 has 𝑎,0 as one of its points
You should recognize this as the 𝑥-intercept of the graph
In FTC2, you will see that our function 𝐹 is really
𝐹 𝑥 =𝐶+ 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
𝐶 is a constant 𝑦-value and the point 𝑎,𝐶 is on the graph of 𝐹

29. The Effect of Changing a in 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
𝐹 𝑥 =𝐶+ 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
𝐶 is a constant 𝑦-value and the point 𝑎,𝐶 is on the graph of 𝐹
Therefore
𝐹 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡=0+ 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
If 𝐶=0, the 𝑎 is an 𝑥-intercept of 𝐹
If you change 𝑎 without changing 𝐶, the graph of 𝐹 changes by shifting vertically

30. The Effect of Changing a in 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡

31. The Fundamental Theorem of Calculus, Part II
If 𝑓 is continuous at every point of [𝑎,𝑏], and if 𝐹 is any antiderivative of 𝑓 on [𝑎,𝑏], then
𝑎 𝑏 𝑓(𝑥) 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎)
This part of the FTC is also called the Integral Evaluation Theorem

32. The Fundamental Theorem of Calculus, Part II
PROOF The FTC1 guarantees the existence of an antiderivative of 𝑓, call it 𝐺 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡 Corollary 3 of the MVT for derivatives says that if 𝐹 is any antiderivative of 𝑓, then 𝐹 𝑥 =𝐺 𝑥 +𝐶, where 𝐶 is a constant. Therefore 𝐹 𝑏 −𝐹 𝑎 = 𝐺 𝑏 +𝐶 − 𝐺 𝑎 +𝐶 𝐹 𝑏 −𝐹 𝑎 =𝐺 𝑏 −𝐺(𝑎)

33. The Fundamental Theorem of Calculus, Part II
PROOF 𝐹 𝑏 −𝐹 𝑎 =𝐺 𝑏 −𝐺(𝑎) 𝐹 𝑏 −𝐹 𝑎 = 𝑎 𝑏 𝑓 𝑡 𝑑𝑡− 𝑎 𝑎 𝑓(𝑡) 𝑑𝑡 𝐹 𝑏 −𝐹 𝑎 = 𝑎 𝑏 𝑓 𝑡 𝑑𝑡−0 𝐹 𝑏 −𝐹 𝑎 = 𝑎 𝑏 𝑓(𝑡) 𝑑𝑡

34. Interpreting FTC2
With FTC2, we can now finally state that a definite integral can be evaluated by finding an antiderivative of the integrand (assuming an antiderivative can be found)
This eliminates the need to rely on calculating Riemann Sums

35. Example 5: Evaluating an Integral
Evaluate −1 3 𝑥 𝑑𝑥 using an antiderivative.

36. Example 5: Evaluating an Integral
Evaluate −1 3 𝑥 𝑑𝑥 using an antiderivative.
−1 3 𝑥 𝑑𝑥= −1 3 𝑥 3 𝑑𝑥+ −1 3 1 𝑑𝑥
= 𝑥 − 𝑥 −1 3 = − − − −1
= − =24

37. Finding Area with the FTC2
Recall that we differentiate the total area of a curve in some interval from the net area of the curve over the interval
The net area is really just another name for the definite integral and its value can be positive, negative, or zero
Total area is always positive
To find total area, you must determine where the graph changes from positive to negative
Then take the integral over the interval for which the graph is above the x-axis and subtract from it the interval for which the graph is below the x-axis

38. Finding Area with the FTC2
Alternatively, you can take the integral of the absolute value of the function
This can be done by rewriting the function as a piecewise function
Then evaluate the integral for each part of the function

39. Example 6: Finding Area with Antiderivatives
Find the area of the region between the curve 𝑦=4− 𝑥 2 , 0≤𝑥≤3, and the x-axis.

40. Example 6: Finding Area with Antiderivatives
Find the area of the region between the curve 𝑦=4− 𝑥 2 , 0≤𝑥≤3, and the x-axis.
This is the graph of an inverted parabola with roots at 𝑥=±2. It is positive over the interval −2<𝑥<2 and negative on 𝑥<−2 and on 𝑥>2. Hence, the net area is positive on [0,2] and negative on [2,3]. The area is
Area= − 𝑥 2 𝑑𝑥− 2 3 (4− 𝑥 2 ) 𝑑𝑥
= 4𝑥− 𝑥 − 4𝑥− 𝑥

41. Example 6: Finding Area with Antiderivatives

42. Finding Area Analytically
To find the area between the graph of 𝑦=𝑓(𝑥) and the x-axis over the interval [𝑎,𝑏] analytically
Partition [𝑎,𝑏] with zeros of 𝑓
Integrate 𝑓 over each subinterval
Add the absolute values of the integrals

43. Example 7: Finding Area Using Your Calculator
Find the area of the region between the curve 𝑦=𝑥 cos 2𝑥 and the x- axis over the interval [−3,3].

44. Example 7: Finding Area Using Your Calculator
Have you calculator find the integral over the given interval for 𝑥 cos 2𝑥 . The result is approximately 5.425

45. Finding Area With a Calculator
To find the area between the graph of 𝑦=𝑓(𝑥) and the x-axis over the interval [𝑎,𝑏] numerically, evaluate
𝑎 𝑏 |𝑓 𝑥 | 𝑑𝑥

46. Analyzing Antiderivative Graphically
A frequent type of free response problem on the AP exam is one where an integral is given (as a graph or a function) and conclusions must be drawn about its antiderivative
For example, suppose that the graph of a function 𝑓 is given and the function ℎ is defined as ℎ(𝑥)= 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
By FTC1 we have that ℎ ′ 𝑥 =𝑓(𝑥)
That is, the graph of 𝑓 is the graph of the derivative of ℎ
Hence, using theorems from Chapter 4, we can draw conclusions about, for example, where ℎ is rising or falling, where maxima or minima are located, and so forth

47. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
The graph of a continuous function 𝑓 with domain [0,8] is shown in the figure. Let ℎ be defined by ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡.
Find ℎ(1).
Is ℎ(0) positive or negative? Justify your answer.
Find the value of 𝑥 for which ℎ(𝑥) is a maximum.
Find the value of 𝑥 for which ℎ(𝑥) is a minimum.
Find the x-coordinate of all points of inflections of the graph of 𝑦= ℎ 𝑥 .

48. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡
ℎ 1 =
ℎ(0) positive? Negative?
ℎ(𝑥) is a maximum
ℎ(𝑥) is a minimum
Points of inflection

49. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
The graph of a continuous function 𝑓 with domain [0,8] is shown in the figure. Let ℎ be defined by ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡.
Find ℎ(1). ℎ 1 =0
Is ℎ(0) positive or negative? Justify your answer. The area over [0,1] is positive, but this means that
ℎ 1 = 1 0 𝑓(𝑡) 𝑑𝑡=− 0 1 𝑓 𝑡 𝑑𝑡 so ℎ 1 <0
Find the value of 𝑥 for which ℎ(𝑥) is a maximum.
Find the value of 𝑥 for which ℎ(𝑥) is a minimum.
Find the x-coordinate of all points of inflections of the graph of 𝑦=ℎ 𝑥 .

50. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
The graph of a continuous function 𝑓 with domain [0,8] is shown in the figure. Let ℎ be defined by ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡.
Find the value of 𝑥 for which ℎ(𝑥) is a maximum. Since the graph of 𝑓 changes from positive to negative at 𝑥=4, then ℎ has a maximum at 𝑥=4

51. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
The graph of a continuous function 𝑓 with domain [0,8] is shown in the figure. Let ℎ be defined by ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡.
Find the value of 𝑥 for which ℎ(𝑥) is a minimum. The sign analysis of the derivative above shows that the minimum value occurs at the endpoint of the interval [0,8]. We see by comparing areas that ℎ 0 = 1 0 𝑓(𝑡) 𝑑𝑡≈−0.5, while ℎ 8 = 1 8 𝑓(𝑡) 𝑑𝑡 is a negative number considerably less than −1. Thus 𝑓(8) is a minimum.

52. Example 8: Using the Graph of 𝑓 to Analyze ℎ 𝑥 = 𝑎 𝑥 𝑓(𝑡) 𝑑𝑡
The graph of a continuous function 𝑓 with domain [0,8] is shown in the figure. Let ℎ be defined by ℎ 𝑥 = 1 𝑥 𝑓(𝑡) 𝑑𝑡.
Find the x-coordinate of all points of inflections of the graph of 𝑦= ℎ(𝑥). The graph of ℎ′(𝑥) changes from decreasing to increasing at 𝑥=1, from increasing to decreasing at 𝑥=3, and from decreasing to increasing at 𝑥=6. These are all points of inflection.