Antiderivatives. Mathematics 11: Lecture 30. Dan Sloughter. Furman University. November 7, 2007

Transcription

1 Antiderivatives Mathematics 11: Lecture 30 Dan Sloughter Furman University November 7, 2007 Dan Sloughter (Furman University) Antiderivatives November 7, / 9

2 Definition Recall: Suppose F and f are defined on an open interval (a, b) with F (x) = f (x) for all x in (a, b). Then we call F an antiderivative of f on (a, b). Dan Sloughter (Furman University) Antiderivatives November 7, / 9

3 Definition Recall: Suppose F and f are defined on an open interval (a, b) with F (x) = f (x) for all x in (a, b). Then we call F an antiderivative of f on (a, b). Recall: If F and G are both antiderivatives of f on an open interval (a, b), then there exists a constant c such that for all x in (a, b). F (x) = G(x) + c Dan Sloughter (Furman University) Antiderivatives November 7, / 9

4 Definition Recall: Suppose F and f are defined on an open interval (a, b) with F (x) = f (x) for all x in (a, b). Then we call F an antiderivative of f on (a, b). Recall: If F and G are both antiderivatives of f on an open interval (a, b), then there exists a constant c such that for all x in (a, b). F (x) = G(x) + c If F is an antiderivative of f, then we call the collection of functions specified by F (x) + c for any constant c is called the general antiderivative of f. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

5 Examples The general antiderivative of f (x) = x 2 is F (x) = 1 3 x 3 + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

6 Examples The general antiderivative of f (x) = x 2 is F (x) = 1 3 x 3 + c. The general antiderivative of f (x) = sin(x) is F (x) = cos(x) + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

7 Properties Using the properties of derivatives, it is clear that if F and G are antiderivatives of f and g, respectively, then Dan Sloughter (Furman University) Antiderivatives November 7, / 9

8 Properties Using the properties of derivatives, it is clear that if F and G are antiderivatives of f and g, respectively, then F (x) + G(x) is an antiderivative of f (x) + g(x), Dan Sloughter (Furman University) Antiderivatives November 7, / 9

9 Properties Using the properties of derivatives, it is clear that if F and G are antiderivatives of f and g, respectively, then F (x) + G(x) is an antiderivative of f (x) + g(x), F (x) G(x) is an antiderivative of f (x) g(x), Dan Sloughter (Furman University) Antiderivatives November 7, / 9

10 Properties Using the properties of derivatives, it is clear that if F and G are antiderivatives of f and g, respectively, then F (x) + G(x) is an antiderivative of f (x) + g(x), F (x) G(x) is an antiderivative of f (x) g(x), kf (x) is an antiderivative of kf (x) for any constant k Dan Sloughter (Furman University) Antiderivatives November 7, / 9

11 Basic antiderivatives Some basic general antiderivatives: Function General Antiderivative x n, n 1 1 n+1 x n+1 + c sin(x) cos(x) + c cos(x) sin(x) + c sec 2 (x) tan(x) + c csc 2 (x) cot(x) + c sec(x) tan(x) sec(x) + c csc(x) cot(x) csc(x) + c Dan Sloughter (Furman University) Antiderivatives November 7, / 9

12 Examples The general antiderivative of f (x) = 4x 3 6x + cos(x) is F (x) = x 4 3x 2 + sin(x) + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

13 Examples The general antiderivative of f (x) = 4x 3 6x + cos(x) is The general antiderivative of F (x) = x 4 3x 2 + sin(x) + c. is f (x) = 3 x sin(x) F (x) = 3 x 8 cos(x) + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

14 Example Suppose f (x) = 3x 4 6x + 1 and f (1) = 4. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

15 Example Suppose f (x) = 3x 4 6x + 1 and f (1) = 4. Then f (x) = 3 5 x 5 3x 2 + x + c for some constant c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

16 Example Suppose f (x) = 3x 4 6x + 1 and f (1) = 4. Then f (x) = 3 5 x 5 3x 2 + x + c for some constant c. And so 4 = f (1) = c = c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

17 Example Suppose f (x) = 3x 4 6x + 1 and f (1) = 4. Then f (x) = 3 5 x 5 3x 2 + x + c for some constant c. And so 4 = f (1) = c = c. Thus and c = = 27 5, f (x) = 3 5 x 5 3x 2 + x Dan Sloughter (Furman University) Antiderivatives November 7, / 9

18 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

19 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Let s be the height of the object after t seconds. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

20 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Let s be the height of the object after t seconds. Then d 2 s dt 2 = 32 feet/second2. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

21 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Let s be the height of the object after t seconds. Then d 2 s dt 2 = 32 feet/second2. So, for some constant c, ds dt = 32t + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

22 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Let s be the height of the object after t seconds. Then d 2 s dt 2 = 32 feet/second2. So, for some constant c, Then ds dt = 32t + c. v 0 = ds dt = c. t=0 Dan Sloughter (Furman University) Antiderivatives November 7, / 9

23 Free fall revisited Suppose an object is thrown vertically into the air from an initial height of s 0 feet with an initial velocity of v 0 feet per second. Let s be the height of the object after t seconds. Then d 2 s dt 2 = 32 feet/second2. So, for some constant c, Then ds dt Hence the velocity of the object is = 32t + c. v 0 = ds dt = c. t=0 v = ds dt = 32t + v 0 feet/second. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

24 Free fall (cont d) Next, for some constant c, s = 16t 2 + v 0 t + c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

25 Free fall (cont d) Next, for some constant c, s = 16t 2 + v 0 t + c. Then s 0 = s t=0 = c. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

26 Free fall (cont d) Next, for some constant c, s = 16t 2 + v 0 t + c. Then So we have s 0 = s t=0 = c. s = 16t 2 + v 0 t + s 0 feet. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

27 Free fall (cont d) Next, for some constant c, s = 16t 2 + v 0 t + c. Then So we have s 0 = s t=0 = c. s = 16t 2 + v 0 t + s 0 feet. Note: If we were working in meters instead of in feet, we would have d 2 s dt 2 = 9.8 meters/second2. Dan Sloughter (Furman University) Antiderivatives November 7, / 9

28 Free fall (cont d) Next, for some constant c, s = 16t 2 + v 0 t + c. Then So we have s 0 = s t=0 = c. s = 16t 2 + v 0 t + s 0 feet. Note: If we were working in meters instead of in feet, we would have d 2 s dt 2 = 9.8 meters/second2. And so s = 4.9t 2 + v 0 t + s 0. Dan Sloughter (Furman University) Antiderivatives November 7, / 9