Antiderivatives! James K. Peterson. January 28, Department of Biological Sciences and Department of Mathematical Sciences Clemson University

Transcription

1 ! James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University January 28, 2014

2 Outline 1 2 Simple Fractional Power

3 Abstract This lecture is going to talk about a thing called antiderivatives which is like the inverse of taking a derivative. We ll work you through all of the details and you ll find it is not so bad. It trains your mind to do stuff in your head which is a good thing!

4 The idea of an Antiderivative or Primitive is very simple. We just guess! We say F is the antiderivative of f if F = f. Since we know a group of simple derivatives, we can guess a group of simple antiderivatives! From this definition, we can see immediately that antiderivatives are not unique. The derivative of any constant is zero, so adding a constant to an antiderivative just gives a new antiderivative!

5 As we said, we can guess many antiderivatives. The symbol we previously introduced as the symbol for a Riemann integral ( be patient, we will be getting to that soon!) is also used to denote the antiderivative.

6 As we said, we can guess many antiderivatives. The symbol we previously introduced as the symbol for a Riemann integral ( be patient, we will be getting to that soon!) is also used to denote the antiderivative. This common symbol for the antiderivative of f has thus evolved to be f because of the close connection between the antiderivative of f and the Riemann integral of f which is given in the Cauchy Fundamental Theorem of Calculus which we will get to in a bit.

7 The usual Riemann integral, b a f (t) dt of f on [a, b] computes a definite value hence, the symbol b a f (t) dt to contrast it with the family of functions represented by the antiderivative f. We will discuss this thing called a Riemann integral shortly. We discussed the idea of it earlier and showed it was a kind of limit.

8 The usual Riemann integral, b a f (t) dt of f on [a, b] computes a definite value hence, the symbol b a f (t) dt to contrast it with the family of functions represented by the antiderivative f. We will discuss this thing called a Riemann integral shortly. We discussed the idea of it earlier and showed it was a kind of limit. Since the antiderivatives are arbitrary up to a constant, most of us refer to the antiderivative as the indefinite integral of f Also, we hardly ever say let s find the antiderivative of f instead, we just say, let s integrate f. We will begin using this shorthand now!

9 Let s begin with antiderivative we can guess for the function t. We know the derivative of t 2 is 2t so it follows the derivative of 1/2 t 2 must be t. In fact, adding a constant doesn t change the result.

10 Let s begin with antiderivative we can guess for the function t. We know the derivative of t 2 is 2t so it follows the derivative of 1/2 t 2 must be t. In fact, adding a constant doesn t change the result. In general, we have for any constant C that d dt ( 1/2t 2 + C ) = t

11 Let s begin with antiderivative we can guess for the function t. We know the derivative of t 2 is 2t so it follows the derivative of 1/2 t 2 must be t. In fact, adding a constant doesn t change the result. In general, we have for any constant C that d dt ( 1/2t 2 + C ) = t We say that 1/2t 2 + C is the family of antiderivatives of t or more simply, just 1/2t 2 + C is the antiderivative of t.

12 Let s begin with antiderivative we can guess for the function t. We know the derivative of t 2 is 2t so it follows the derivative of 1/2 t 2 must be t. In fact, adding a constant doesn t change the result. In general, we have for any constant C that d dt ( 1/2t 2 + C ) = t We say that 1/2t 2 + C is the family of antiderivatives of t or more simply, just 1/2t 2 + C is the antiderivative of t. Note we could also say these are the primitives of t too.

13 Let s begin with antiderivative we can guess for the function t. We know the derivative of t 2 is 2t so it follows the derivative of 1/2 t 2 must be t. In fact, adding a constant doesn t change the result. In general, we have for any constant C that d dt ( 1/2t 2 + C ) = t We say that 1/2t 2 + C is the family of antiderivatives of t or more simply, just 1/2t 2 + C is the antiderivative of t. Note we could also say these are the primitives of t too. The usual symbol for the antiderivative is the Riemann integral symbol without the a and b. So we would say t dt represents the antiderivative of t and t dt = 1/2t 2 + C.

14 Next, let s look at the function t 2. We know the derivative of t 3 is 3t 2 so it follows the derivative of 1/3 t 3 must be t 2. In fact, adding a constant doesn t change the result.

15 Next, let s look at the function t 2. We know the derivative of t 3 is 3t 2 so it follows the derivative of 1/3 t 3 must be t 2. In fact, adding a constant doesn t change the result. In general, we have for any constant C that ( ) d 1/3t 3 + C = t 2 dt

16 Next, let s look at the function t 2. We know the derivative of t 3 is 3t 2 so it follows the derivative of 1/3 t 3 must be t 2. In fact, adding a constant doesn t change the result. In general, we have for any constant C that ( ) d 1/3t 3 + C = t 2 dt We say that 1/3t 3 + C is the family of antiderivatives of t 2 or more simply, just 1/3t 3 + C is the antiderivative of t 2.

17 Next, let s look at the function t 2. We know the derivative of t 3 is 3t 2 so it follows the derivative of 1/3 t 3 must be t 2. In fact, adding a constant doesn t change the result. In general, we have for any constant C that ( ) d 1/3t 3 + C = t 2 dt We say that 1/3t 3 + C is the family of antiderivatives of t 2 or more simply, just 1/3t 3 + C is the antiderivative of t 2. Note we could also say these are the primitives of t 2 too.

18 Next, let s look at the function t 2. We know the derivative of t 3 is 3t 2 so it follows the derivative of 1/3 t 3 must be t 2. In fact, adding a constant doesn t change the result. In general, we have for any constant C that ( ) d 1/3t 3 + C = t 2 dt We say that 1/3t 3 + C is the family of antiderivatives of t 2 or more simply, just 1/3t 3 + C is the antiderivative of t 2. Note we could also say these are the primitives of t 2 too. Also, we would say for any constant C that t 2 dt = 1/3t 3 + C.

19 We can do a similar analysis for other powers. You should be able to convince yourself that for these positive powers, we have 1 dt = t + C.

20 We can do a similar analysis for other powers. You should be able to convince yourself that for these positive powers, we have 1 dt = t + C. t dt = 1/2 t 2 + C.

21 We can do a similar analysis for other powers. You should be able to convince yourself that for these positive powers, we have 1 dt = t + C. t dt = 1/2 t 2 + C. t 2 dt = 1/3 t 3 + C.

22 We can do a similar analysis for other powers. You should be able to convince yourself that for these positive powers, we have 1 dt = t + C. t dt = 1/2 t 2 + C. t 2 dt = 1/3 t 3 + C. t 3 dt = 1/4 t 4 + C.

23 Further, we can guess for negative powers also. We can do a similar analysis for negative powers. You should be able to convince yourself that t 2 dt = t 1 + C.

24 Further, we can guess for negative powers also. We can do a similar analysis for negative powers. You should be able to convince yourself that t 2 dt = t 1 + C. t 3 dt = 1/2t 2 + C.

25 Further, we can guess for negative powers also. We can do a similar analysis for negative powers. You should be able to convince yourself that t 2 dt = t 1 + C. t 3 dt = 1/2t 2 + C. t 4 dt = 1/3 t 3 + C.

26 Further, we can guess for negative powers also. We can do a similar analysis for negative powers. You should be able to convince yourself that t 2 dt = t 1 + C. t 3 dt = 1/2t 2 + C. t 4 dt = 1/3 t 3 + C. t 5 dt = 1/4 t 4 + C.

27 Further, we can guess for negative powers also. We can do a similar analysis for negative powers. You should be able to convince yourself that t 2 dt = t 1 + C. t 3 dt = 1/2t 2 + C. t 4 dt = 1/3 t 3 + C. t 5 dt = 1/4 t 4 + C. We can then glue together these antiderivatives to handle polynomials!

28 Example Example Find (2t + 3) dt. Solution (2t + 3) dt = (2t) dt + 3 dt = 2 t dt dt = t 2 + 3t + C where the C indicates that we can add any constant we want and still get an antiderivative. C is often called the integration constant.

29 Example Example Find (5t 2 + 8t 2) dt. Solution ( ) ( ) ( ) (5t 2 + 8t 2) dt = 5 t 3 /3 + 8 t 2 /2 2 t + C.

30 Example Example Find (5t 3 + 8t 2 2t) dt. Solution ( ) (5t 3 + 8t 2 2t) dt = 5 t 2 /( 2) ( ) 8 t 1 /( 1) + ( ) 2 t 2 /2 + C

31 Example Example Find (5t 5 + 4t ) dt. Solution ( ) ( ) (5t 5 + 4t ) dt = 5 t 6 /6 + 4 t 3 /3 + 20t + C.

32 Homework Find (15t 4 + 4t 3 + 9t + 7) dt Find (6t 3 4t 2 12) dt Find (50t t 2/(t 2 )) dt Find (12 + 8t 7 2t 12 ) dt Find (4 + 7t) dt Find (6 + 3t 4 ) dt Find (1 + 2t + 3t 2 ) dt Find ( t 4t 5 ) dt.

33 Simple Fractional Power Now we haven t yet discussed derivative of fractional powers of x. It is not that hard but it is easy to get blown away by a listing of too many mathy things, boom, one after the other. Here is a simple example to show you how to do it.

34 Simple Fractional Power Now we haven t yet discussed derivative of fractional powers of x. It is not that hard but it is easy to get blown away by a listing of too many mathy things, boom, one after the other. Here is a simple example to show you how to do it. Consider f (x) = x 2/3. For convenience, let y = f (x). Then we have y = x 2/3. Cube both sides to get y 3 = x 2.

35 Simple Fractional Power Now we haven t yet discussed derivative of fractional powers of x. It is not that hard but it is easy to get blown away by a listing of too many mathy things, boom, one after the other. Here is a simple example to show you how to do it. Consider f (x) = x 2/3. For convenience, let y = f (x). Then we have y = x 2/3. Cube both sides to get y 3 = x 2. Now use the chain rule on the left hand side and a regular derivative on the right hand side to get 3 y 2 y = 2 x. Now we just manipulate 3 y 2 y = 2 x = y = 2 x 3 y 2.

36 Simple Fractional Power Now we haven t yet discussed derivative of fractional powers of x. It is not that hard but it is easy to get blown away by a listing of too many mathy things, boom, one after the other. Here is a simple example to show you how to do it. Consider f (x) = x 2/3. For convenience, let y = f (x). Then we have y = x 2/3. Cube both sides to get y 3 = x 2. Now use the chain rule on the left hand side and a regular derivative on the right hand side to get 3 y 2 y = 2 x. Now we just manipulate 3 y 2 y = 2 x = y = 2 x 3 y 2. But, we can plug in for y 2 = x 4/3 to get y = 2 x 3 x 4/3 = 2 3 x 1 4/3 = 2 3 x 1/3.

37 Simple Fractional Power This mix of chain rule and regular differentiation is an easy trick. You can see we can do this for any fraction p/q. We get another theorem! Theorem The Simple Power Rule: Fractions! If f is the function x p/q for any integer p and q except q = 0, of course, then the derivative of f with respect to x satisfies (x p/q) = p q x p/q 1 Proof We did the example for the power 2/3 but the reasoning is the same for any fraction!

38 Simple Fractional Power So we can also find antiderivatives of fractional powers. You should be able to convince yourself that t 1 2 dt = 2/3 t 3/2 + C.

39 Simple Fractional Power So we can also find antiderivatives of fractional powers. You should be able to convince yourself that t 1 2 dt = 2/3 t 3/2 + C. t 4 5 dt = 5/9 t 9/5 + C.

40 Simple Fractional Power So we can also find antiderivatives of fractional powers. You should be able to convince yourself that t 1 2 dt = 2/3 t 3/2 + C. t 4 5 dt = 5/9 t 9/5 + C. t 5 8 dt = 8/13 t 13/8 + C.

41 Simple Fractional Power So we can also find antiderivatives of fractional powers. You should be able to convince yourself that t 1 2 dt = 2/3 t 3/2 + C. t 4 5 dt = 5/9 t 9/5 + C. t 5 8 dt = 8/13 t 13/8 + C. t 2 3 dt = 3 t 1/3 + C.

42 Simple Fractional Power It is thus easy to guess the antiderivative of a power of t as we have already mentioned. Here is the Theorem. Theorem Of Simple Fractional Powers If u is any fractional power other than 1, then the antiderivative of f (t) = t u is F (t) = t u+1 /(u + 1) + C. This is also expressed as t u dt = t u+1 /(u + 1) + C. Since this result holds for any fractional power and fractions and irrational numbers can t be isolated from one another, we can show more. The rule above holds for any number other than 1: even a number like 2. But we won t belabor this point now. Proof This is just a statement of all the results we have gone over.

43 Simple Fractional Power Example Example Find t 4/5 dt. Solution t 4/5 dt = t 9/5 /(9/5) + C.

44 Simple Fractional Power Example Example Find (t 1/2 + 9t 1/3 ) dt. Solution ( ) (t 1/2 + 9t 1/3 ) dt = t 3/2 /(3/2) + 9 t 4/3 /(4/3) + C.

45 Simple Fractional Power Example Example Find (6t 5/7 ) dt. Solution ( ) (6t 5/7 ) dt = 6 t 12/7 /(12/7) + C.

46 Simple Fractional Power Example Example Find (8t 3/4 + 12t 1/5 ) dt. Solution ( ) (8t 3/4 + 12t 1/5 ) dt = 8 t 1/4 /(1/4) ( ) +12 t 4/5 /(4/5) + C.

47 Simple Fractional Power Homework Find (6 t 2/7 ) dt Find (4 t 3/2 + 5t 1/3 ) dt Find (20 t 12/5 ) dt Find (2 t 2/7 + 14t 5/8 ) dt Find ( 22 t 11/3 + 6 t 1/4 ) dt Find (3 x 9/8 ) dx Find (6 u 4/3 + 5 u 7/2 ) du Find ( 19y 1/6 ) dy.