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4.9 – Antiderivatives.

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1 4.9 – Antiderivatives

2 In addition to finding derivatives, there is an important β€œinverse” problem
Given the derivative, find the function itself. For example, in physics we may know the velocity 𝑣 𝑑 (the derivative) and wish to compute the position 𝑠 𝑑 of an object. Since 𝑠 β€² 𝑑 =𝑣 𝑑 , this amounts to finding a function whose derivative is 𝑣 𝑑 . A function 𝐹 π‘₯ whose derivative is 𝑓 π‘₯ is called an antiderivative of 𝑓 π‘₯ .

3 Definition Antiderivatives
A function 𝐹 π‘₯ is an antiderivative of 𝑓 π‘₯ on π‘Ž, 𝑏 if 𝐹 β€² π‘₯ =𝑓 π‘₯ for all π‘₯∈ π‘Ž, 𝑏 .

4 Examples 𝐹 π‘₯ =βˆ’ cos π‘₯ is an antiderivative of 𝑓 π‘₯ = sin π‘₯ because 𝐹 β€² π‘₯ = 𝑑 𝑑π‘₯ βˆ’ cos π‘₯ = sin π‘₯ =𝑓 π‘₯ 𝐹 π‘₯ = 1 3 π‘₯ 3 is an antiderivative of 𝑓 π‘₯ = π‘₯ 2 because 𝐹 β€² π‘₯ = 𝑑 𝑑π‘₯ π‘₯ 3 = π‘₯ 2 =𝑓 π‘₯

5 One critical observation is that antiderivatives are not unique.
We are free to add a constant C because the derivative of a constant is zero. If 𝐹 β€² π‘₯ =𝑓 π‘₯ , then 𝐹 π‘₯ +𝐢 β€² =𝑓 π‘₯ . Each of the following is an antiderivative of π‘₯ 2 1 3 π‘₯ 3 1 3 π‘₯ 3 +5 1 3 π‘₯ 3 βˆ’4 The process of finding an antiderivative is called integration.

6 Notation Indefinite Integral
The notation 𝑓 π‘₯ 𝑑π‘₯ =𝐹 π‘₯ +𝐢 means that 𝐹 β€² π‘₯ =𝑓 π‘₯ We say that 𝐹 π‘₯ +𝐢 is the general antiderivative or indefinite integral of 𝑓 π‘₯

7 Power Rule for Integrals

8 Antiderivative of y = 1/x
The function 𝐹 π‘₯ = ln π‘₯ is an antiderivative of 𝑦= 1 π‘₯ in the domain π‘₯:π‘₯β‰ 0 ; that is 𝑑π‘₯ π‘₯ = ln π‘₯ +𝐢

9 Linearity of the Indefinite Integral
Sum Rule 𝑓 π‘₯ +𝑔 π‘₯ 𝑑π‘₯ = 𝑓 π‘₯ 𝑑π‘₯ + 𝑔 π‘₯ 𝑑π‘₯ Multiples Rule 𝑐𝑓 π‘₯ 𝑑π‘₯ =𝑐 𝑓 π‘₯ 𝑑π‘₯

10 Basic Trigonometric Integrals
sin π‘₯ 𝑑π‘₯ =βˆ’ cos π‘₯ +𝐢 cos π‘₯ 𝑑π‘₯ = sin π‘₯ +𝐢 sec 2 π‘₯ 𝑑π‘₯ = tan π‘₯ +𝐢 csc 2 π‘₯ 𝑑π‘₯ =βˆ’ cot π‘₯ +𝐢 sec π‘₯ tan π‘₯ 𝑑π‘₯ = sec π‘₯ +𝐢 csc π‘₯ cot π‘₯ 𝑑π‘₯ =βˆ’ csc π‘₯ +𝐢

11 Integrals Involving ex
𝑒 π‘₯ 𝑑π‘₯ = 𝑒 π‘₯ +𝐢 𝑒 π‘˜π‘₯+𝑏 𝑑π‘₯ = 1 π‘˜ 𝑒 π‘˜π‘₯+𝑏 +𝐢

12 Examples Evaluate cos π‘₯ 𝑑π‘₯ Evaluate 3 π‘₯ 4 βˆ’5 π‘₯ 2/3 + π‘₯ βˆ’3 𝑑π‘₯

13 Examples Evaluate sin 8π‘‘βˆ’3 +20 cos 9𝑑 𝑑𝑑 Evaluate 3 𝑒 π‘₯ βˆ’4 𝑑π‘₯

14 We can think of an antiderivative as a solution to the differential equation 𝑑𝑦 𝑑π‘₯ =𝑓 π‘₯
In general, a differential equation is an equation relating an unknown function and its derivatives. A differential equation with an initial condition is called an initial value problem.

15 Examples Solve 𝑑𝑦 𝑑π‘₯ =4 π‘₯ 7 with an initial condition 𝑦 0 =4.

16 Examples Solve the differential equation 𝑑𝑦 𝑑𝑑 = sin πœ‹π‘‘ with the initial condition 𝑦 2 =2

17 Homework Text page 281 #s even, even, all

18 Speed Quiz

19 5.3 – The Fundamental Theorem of Calculus, Part I

20 Fundamental Theorem of Calculus Part I
Assume that 𝑓 π‘₯ is continuous on π‘Ž, 𝑏 . If 𝐹 π‘₯ is an antiderivative of 𝑓 π‘₯ on π‘Ž, 𝑏 , then π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ =𝐹 𝑏 βˆ’πΉ π‘Ž OR π‘Ž 𝑏 𝑓 β€² π‘₯ 𝑑π‘₯ =𝑓 𝑏 βˆ’π‘“ π‘Ž

21 Other Important Integral Theorems
π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = signed area of region between the graph and x-axis over [a, b] For a < b, 𝑏 π‘Ž 𝑓 π‘₯ 𝑑π‘₯ =βˆ’ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ π‘Ž π‘Ž 𝑓 π‘₯ 𝑑π‘₯ =0 Let π‘Žβ‰€π‘β‰€π‘, and assume that 𝑓 π‘₯ is integrable. π‘Ž 𝑐 𝑓 π‘₯ 𝑑π‘₯= π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ + 𝑏 𝑐 𝑓 π‘₯ 𝑑π‘₯ π‘Ž 𝑏 𝑑π‘₯ π‘₯ = ln 𝑏 βˆ’ ln π‘Ž = ln 𝑏 π‘Ž For an object in linear motion with velocity 𝑣 𝑑 , then Displacement during 𝑑 1 , 𝑑 2 = 𝑑1 𝑑2 𝑣 𝑑 𝑑𝑑 Distance traveled during 𝑑 1 , 𝑑 2 = 𝑑1 𝑑2 𝑣 𝑑 𝑑𝑑

22 Example Calculate the area under the graph of 𝑓 π‘₯ = π‘₯ 3 over 2, 4

23 Example Calculate βˆ’πœ‹/4 πœ‹/4 sec 2 π‘₯ 𝑑π‘₯ and sketch the corresponding region.

24 Example Calculate 0 πœ‹ sin π‘₯ 𝑑π‘₯ and sketch the corresponding region.

25 Example Calculate 0 2πœ‹ sin π‘₯ 𝑑π‘₯ and sketch the corresponding region.

26 Example Calculate 0 2 𝑓 π‘₯ 𝑑π‘₯ given 𝑓 π‘₯ = 3 π‘₯ for π‘₯<1 2π‘₯+1 for π‘₯β‰₯1

27 Example Calculate βˆ’2 5 𝑓 π‘₯ 𝑑π‘₯ given 𝑓 π‘₯ = 2π‘₯βˆ’ π‘₯ 2 for π‘₯< for π‘₯β‰₯1

28 Example Calculate βˆ’3 2 π‘₯ 𝑑π‘₯ by first rewriting the integral without absolute value.

29 Example Calculate βˆ’1 4 π‘₯βˆ’2 𝑑π‘₯ by first rewriting the integral without absolute value.

30 Homework Text pages #s 6-14 even, 28, 30, 38, 43, 44, 55, 56

31 5.4 – The Fundamental Theorem of Calculus, Part II

32 Example Find a formula for the area function 𝐴 π‘₯ = 3 π‘₯ 𝑑 2 𝑑𝑑 .

33 Example Find the formula for the area function 𝐴 π‘₯ = 2 π‘₯ 2 2π‘₯+5 𝑑π‘₯

34 Fundamental Theorem of Calculus Part II
Assume that 𝑓 π‘₯ is continuous on an open interval I and let π‘ŽβˆˆπΌ. Then the area function 𝐴 π‘₯ = π‘Ž π‘₯ 𝑓 𝑑 𝑑𝑑 is an antiderivative of 𝑓 π‘₯ on I; that is, 𝐴 β€² π‘₯ =𝑓 π‘₯ . Equivalently, 𝑑 𝑑π‘₯ π‘Ž π‘₯ 𝑓 𝑑 𝑑𝑑 =𝑓 π‘₯ Furthermore, 𝐴 π‘₯ satisfies the initial condition 𝐴 π‘Ž =0 If the upper bound is a function of x, apply the chain rule and multiply by the derivative of the upper bound.

35 Example Find the derivative of 𝐴 π‘₯ = 2 π‘₯ 1+ 𝑑 3 𝑑𝑑 . Evalute 𝐴 β€² 2 , 𝐴 β€² 3 , and 𝐴 2 .

36 Example Find the derivative of 𝐺 π‘₯ = βˆ’2 π‘₯ 2 sin 𝑑 𝑑𝑑

37 Homework Text page 320 #s 4-10 even, 28

38 Quiz

39 5.6 – Substitution Method

40 Integration (antidifferentiation) is generally more difficult than differentiation.
There are no sure-fire methods, and many antiderivatives cannot be expressed in terms of elementary functions. However, there are a few important general techniques. One such technique is the Substitution Method, which uses the Chain Rule β€œin reverse.” Consider the integral 2π‘₯ cos π‘₯ 2 𝑑π‘₯ . We can evaluate it if we remember the Chain Rule 𝑑 𝑑π‘₯ sin π‘₯ 2 =2π‘₯ cos π‘₯ 2 This tells us that sin π‘₯ 2 is an antiderivative of 2π‘₯ cos π‘₯ 2 , and therefore 2π‘₯ Derivative of the inside function cos π‘₯ 2 Inside function 𝑑π‘₯=sin⁑ π‘₯ 2 +𝐢

41 The integrand is the product of a composite function and the derivative of the inside function.
The Chain Rule does not help if the derivative of the inside function is missing. For instance, we cannot use the Chain Rule to compute cos π‘₯+ π‘₯ 3 𝑑π‘₯ because the factor 1+3 π‘₯ 2 does not appear.

42 Theorem – The Substitution Method
If 𝐹 β€² π‘₯ =𝑓 π‘₯ , then 𝑓 𝑒 π‘₯ 𝑒 β€² π‘₯ 𝑑π‘₯ =𝐹 𝑒 π‘₯ +𝐢

43 Equivalently, 𝑑𝑒 and 𝑑π‘₯ are related by 𝑑𝑒= 𝑒 β€² π‘₯ 𝑑π‘₯ For example
Before proceeding to the examples, we discuss the procedure for carrying out substitution using differentials. Differentials are symbols such as 𝑑𝑒 or 𝑑π‘₯ that occur in the Leibniz notations. In our calculators, we shall manipulate them as thorugh they are related by an equation in which the 𝑑π‘₯ β€œcancels”: 𝑑𝑒= 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Equivalently, 𝑑𝑒 and 𝑑π‘₯ are related by 𝑑𝑒= 𝑒 β€² π‘₯ 𝑑π‘₯ For example If 𝑒= π‘₯ 2 then 𝑑𝑒=2π‘₯𝑑π‘₯ If 𝑒= cos π‘₯ 3 then 𝑑𝑒=βˆ’3 π‘₯ 2 sin π‘₯ 3 𝑑π‘₯

44 Example Evalaute 3 π‘₯ 2 sin π‘₯ 3 𝑑π‘₯

45 Example Evalaute 3 π‘₯ 2 sin π‘₯ 3 𝑑π‘₯

46 Example Evalaute π‘₯ π‘₯ 𝑑π‘₯

47 Example Evaluate π‘₯ 2 +2π‘₯ 𝑑π‘₯ π‘₯ 3 +3 π‘₯

48 Example Evalaute sin 7πœƒ+5 π‘‘πœƒ

49 Example Evaluate 𝑒 βˆ’9𝑑 𝑑𝑑

50 Change of Variables for Definite Integrals
π‘Ž 𝑏 𝑓 𝑒 π‘₯ 𝑒 β€² π‘₯ 𝑑π‘₯ = 𝑒 π‘Ž 𝑒 𝑏 𝑓 𝑒 𝑑𝑒

51 Example Evaluate π‘₯ 2 π‘₯ 3 +1 𝑑π‘₯

52 example Evaluate π‘₯+1 π‘₯ 2 +2π‘₯ 5 𝑑π‘₯

53 example Evalaute 0 πœ‹/4 tan 3 πœƒ sec 2 πœƒ π‘‘πœƒ

54 example Calculate the area under the graph of 𝑦= π‘₯ π‘₯ over 1, 3

55 Homework Text pages #s even, 48, 50, 80, 82, 86


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