AP Calculus BC : The Fundamental Theorem of Calculus

Transcription

1 AP Calculus BC : The Fundamental Theorem of Calculus Tuesday, November 5, 008

2 Homework Answers 6. (a) approimately 0.5 (b) approimately 1 (c) approimately

3 Introduction In this lesson, we introduce the Fundamental Theorem of Calculus, a result that establishes a relationship between antiderivatives and definite integrals. It will be the basis for much of the remainder of this course.

4 Definition Recall that for a function f( ), the average value f( ) of the function on an interval [ ab, ] is given by ( ) 1 b f = f( ) d. This evokes the analogy to b aa finding the height of a rectangle whose width is the interval s width such that the rectangle s area is the same as the function s definite integral on that interval.

5 Eercise 1 Find the average value of f( ) = 5 on [ 5,5 ].

6 Eercise 1 Find the average value of f( ) = 5 on [ 5,5 ]. 5 5 f( ) = 1 = 1 5 d= 5 d 5 ( 5) We can evaluate this integral by noting that it gives the area of half a circle with radius 5, so its value is () 5 5 ( ) 1 1π 5 = π. So f = =

7 Proposition The Fundamental Theorem of Calculus, part I For a real-valued function f continuous on [ ab, ] and for F ( ) = ftdt ( ), the theorem states that a d( ( )) ( ), or equivalently, d F f f() = t dt, f( ) =. d d a

8 Eercise Use the limit derivative definition to show that the Fundamental Theorem of Calculus, part I is true.

9 Eercise Use the limit derivative definition to show that the Fundamental Theorem of Calculus, part I is true. We take and + hto be in ( ab, ) and h 0. Here, + h F ( + h) = ftdt () and F ( ) = ftdt (), so a ( ) ( ) () () F+ h F= ftdt ftdt. We split up the left integral to get a + h ( ) ( ) () () () F+ h F= ftdt+ ftdt ftdt. a + h a a a

10 Eercise Use the limit derivative definition to show that the Fundamental Theorem of Calculus, part I is true. Clearly this is + h ( ) ( ) ( ) F+ h F= ftdt. We ( ) ( ) 1 h divide by hto get F+ h F + = f() t dt. h h Taking the limit as h 0, we have lim F+ h F h 0 h 1 + h = lim f() t dt. h 0 h ( ) ( )

11 Eercise Use the limit derivative definition to show that the Fundamental Theorem of Calculus, part I is true. The left side is d ( F ( )) by definition, and the right d side is the limit of the average value of f() t on [, + h] as h 0, which is clearly just f( ). d Therefore ( F ( )) = f ( ). d

12 Eercise 3 Use the Fundamental Theorem of Calculus, part I to find F ( ) for each of the following. (a) F ( ) = tdt (d) F ( ) = 0 5 costdt (b) F ( ) = 3 sin tdt (e) F ( ) = 3 t edt (c) F ( ) = 3 sin tdt

13 Eercise 3 Use the Fundamental Theorem of Calculus, part I to find F ( ) for each of the following. (a) F ( ) = (d) F ( ) = cos( ) (b) F ( ) = sin (e) = F ( ) e e 3 (c) F ( ) = sin

14 Eercise 4 Find each indefinite integral in the most general form. (a) 3 d (d) 1 1+ d (b) cosd (e) e d (c) 3 d

15 Eercise 4 Find each indefinite integral in the most general form. (a) 3 + A (d) arctan+ D (b) sin+ B (e) 1 e + E (c) 3ln + C (for any A, B, C, D, and E R)

16 Proposition The Fundamental Theorem of Calculus, part II For a real-valued function f continuous on [ ab, ] and for F any antiderivative of f, the theorem states that b f( ) d= F( b) F( a). a

17 Eercise 5 Use (i) the Fundamental Theorem of Calculus, part II and (ii) your calculator to evaluate the following definite integrals. (a) 0 ( 4 ) d (b) 0 π sind

18 Eercise 5 Use (i) the Fundamental Theorem of Calculus, part II and (ii) your calculator to evaluate the following definite integrals. 3 π (a) (b) cos] 0= 1 ( 1) = 3 = 0

19 Eercise 6 Find dy d for each of the following functions. (a) y= e t dt 3 (c) 3 y = 4 ln tdt (b) 3 y = tsintdt (d) y= 4 t 1 + t dt

20 Eercise 6 Find dy d for each of the following functions. (a) y = e (c) y = 4ln 4 (b) y = sin( ) 3 = sin( ) (d) y =

21 Eercise 7 a Find a if ( 6 ) d= 4. 0

22 Eercise 7 a Find a if ( 6 ) d= 4. 0 ( ) 3 6 d= C, so we wish to find a 3 a such that 3 = 4. This is 3a a = 4 which solves to a { 1.091, 1.44, 8,847 }.

23 Eercise 8 Find the area inside the parabola its verte to the line =4. = ( y ) from

24 Eercise 8 Find the area inside the parabola its verte to the line =4. = ( y ) from The parabola s verte is at y=, 0,. We rewrite the equation as y=±, and the area in question is the same as twice the area bounded below by the -ais and above by y= for from 0 to 4. 4 Thus we have ( 3 A= d= ) 4 3 = ( ) ( )

25 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments, as shown below. Let g be the function given by g ( ) ftdt ( ). = 0

26 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (a) Find g( 3. )

27 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (a) Find g( 3. ) This is the signed area bounded by the graph and the -ais for from 0 to 3, which is one fourth of a circle of radius minus a triangle of area 1. That is A= π( ) 1 = π

28 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (b) Find all (,5) at which g has a local maimum. Justify your answer.

29 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (b) Find all (,5) at which g has a local maimum. Justify your answer. For the points to represent a local maimum, we also require that g ( ) changes from positive to negative. The only such point is =.

30 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (c) Write an equation for the line tangent to the graph of y= g( ) at = 3.

31 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (c) Write an equation for the line tangent to the graph of y= g( ) at = 3. The tangent line s slope here is g ( 3) = f( 3) = 1; we have already found g ( 3) = π 1, so in point- slope form the line is y π 1 = 3. ( ) ( )

32 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (d) Find the -coordinate of each point of inflection of the graph of y= g( ) on the open interval (,5 ). Justify your answer.

33 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (d) Find the -coordinate of each point of inflection of the graph of y= g( ) on the open interval (,5 ). Justify your answer. A function s points of inflection occur at the input coordinates at which the second derivative changes sign; in this case, the point(s) of inflection of g ( ) are those for which g ( ) = f ( ) changes sign.

34 Eercise 9 (1997 AB5 BC5) The graph of a function f consists of a semicircle and two line segments. Let g ( ) = ftdt ( ). 0 (d) Find the -coordinate of each point of inflection of the graph of y= g( ) on the open interval (,5 ). Justify your answer. Note that f ( ) is positive just to the left of = 0 and is negative just to its right; also, f ( ) is negative just to the left of = 3 and positive just to its right. Therefore, the points of inflection occur at { 0,3 }.

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