Section 4.4 The Fundamental Theorem of Calculus

Section 4.4 The Fundamental Theorem of Calculus

The First Fundamental Theorem of Calculus If f is continuous on the interval [a,b] and F is any function that satisfies F '() = f() throughout this interval then b f d F b F a a Alternative forms: b F b F a f d ' a f d f b f a b a

Eample Evaluate d 0 First Find the indefinte integral F(): F d C C Now apply the FTC to find the definite integral: F b F a F F0 C 0 C CC Notice that it is not necessary to include the C with definite integrals

Eample Evaluate d First Find the indefinte integral F(): F d C C 4 4 Now apply the FTC to find the definite integral: F b F a F F 4 4 4 C 4 C 7 8 CC Notice that it is 9 4 4 not necessary to include the C with definite integrals

More Eamples: New Notation F() 4. Evaluate sin 5. Evaluate cos d sin sin Bounds 5 0 d 5 0 d If needed, rewrite. 5 0 9 9 6. Evaluate d 5 88 5 05 0 9 4 4 9 9 4 4 4.5 4 d

Eample 7 Calculate the total area between the curve y = and the -ais over the interval [0,]. The question considers all area to be positive (not signed area), thus use the absolute value function:,,, Use a integral and the piece-wise function to find the area: d 0 d d 0 0 0 0 Rewrite the equation as a piece-wise function. 4

Eample 8 Assume F '() = f (), f () = sin ( ), and F() = -5. Find F (). Use the First Fundamental sin d F F Theorem of Calculus: We do not have the ability to analytically calculate this integral. It will either be given or you can use a calculator to evaluate the integral. d F sin 5 0.495 5 F F 5.495

Eample 9 The graph below is of the function f '(). If f (4) =, find f (). Use the First Fundamental Theorem of Calculus: 4 f ' d f f 4 4 f f 8

White Board Challenge If, for all, f '() = ( ) 4 ( ), it follows that the function f has: a) a relative minimum at =. b) a relative maimum at =. c) both a relative minimum at = and a relative maimum at =. d) neither a relative maimum nor a relative minimum. e) relative minima at = and at =. Multiple Choice

Mean Value Theorem Let F f be a function that satisfies the following hypotheses:. F f is is continuous on the closed interval [a,b]. F f is is differentiable on the open interval (a,b) Then there is a number c in (a,b) such that: Ff b Ff a c ff' c b a Redefine the Conditions f c b f d a ba Rewrite with integral notation. b a f c f d b a Solve for the integral.

Mean Value Theorem for Integrals If f() is continuous on [a,b], then there eists a value c on the interval [a,b] such that: b a f d b a f c

Average Value of a Function The average value of an integrable function f() on [a,b] is the quantity: b b a a f d This is also referred to as the Mean Value and can be described as the average height of a graph.

Reminder: Average Rate of Change For a b, the average rate of change of f over time [a,b] is the ratio: f b f a ba Approimates the derivative of a function.

Eample Find the average value of f () = sin on [0,π]. Use the Formula: sin d 0 0 cos 0 cos cos 0

Eample The height of a jump of a bushbaby is modeled by h (t) = v 0 t ½gt. If g = 980 cm/s and the initial velocity is v 0 =600 cm/s, find the average speed during the jump. Use the Average Value Formula: b b h t v t gt a Velocity is the derivative of Position. f d a 0 600 980 600 490 h t t t h t t t Thus, the speed function is: v t h ' t 600 980t We are trying to find the average value of SPEED (absolute value of velocity). So we need to find the velocity function. Now find when the jump begins and ends (a and b). 0 600t490t t 0, 60 49 Evaluate the integral: 0 60 0 49 60 49 600 980t dt

Eample (Continued) Rewrite the equation as a piece-wise function: 600 980t The height of a jump of a bushbaby is modeled by h (t) = v 0 t ½gt. If g = 980 cm/s and the initial velocity is v 0 =600 cm/s, find the average speed during the jump. 60 0 49 600 980 t, 0 t 600 980 t, t 0 60 49 49 Use a integral and the piece-wise function to find the average value: 60 0 60 49 600 980t dt 49 49 49 60 600 980t dt 600 980t dt 0 0 49 0 60 49 49 49 60 600t 490t 600t 490t 0 0 49 600 490 600 0 490 0 600 490 600 490 49 0 0 60 60 0 0 60 49 49 49 49 49 49 49 8000 60 49 00 cm / s 0 49

White Board Challenge Evaluate lim h0 cos h h

Net Change of a Quantity over a Specified Interval Consider the following problems:. Water flows into an empty bucket at a rate of.5 liters/second. How much water is in the bucket after 4 seconds? Quantity of water flow rate time elapsed.54 6 liters. Suppose the flow rate varies with time and can be represented as r(t). How much water is in the bucket after 4 seconds? The quantity of water is equal to the area under the curve of r(t)

Net Change of a Quantity over a Specified Interval Water flows into an empty bucket at a rate of.5 liters/second. Suppose the flow rate varies with time and can be represented as r(t). How much water is in the bucket after 4 seconds? The quantity of water is equal to the area under the curve of r(t). Let s(t) be the amount of water in the bucket at time t. Use the First Fundamental Theorem of Calculus: 0 4 0 s ' t dt s 4 s 0 4 r t dt Water in the bucket at 4 s s 4 s 4 0 4 IMPORTANT: If the bucket did not start empty, the integral would represent the net change of water. r t dt 0 Signed Area under the graph

Net Change as the Integral of a Rate The net change in s(t) over an interval [t,t ] is given by the integral: Rate at which s(t) is changing t t Amount of the quantity at t s ' t dt s t s t Integral of the rate of change Net change from t to t

Eample If b(t) is the rate of growth of the number of bacteria in a dish measured in number of bacteria per hour, what does the following integral represent? Be specific. c a b t dt The increase in the number of bacteria from hour a to hour c.

Eample The number of cars per hour passing an observation point along a highway is called the traffic flow rate q(t) (in cars per hour). The flow rate is recorded in the table below. Estimate the number of cars using the highway during this -hour period. t 7:00 7:5 7:0 7:45 8:00 8:5 8:0 8:45 9:00 q(t) 044 97 478 844 45 78 55 80 54 Since there is no function, we can not use the First Fundamental Theorem of Calculus. Instead approimate the area under the curve with any Riemann Sum (I will use right-endpoints with 0.5 hour lengths): 9:00 q t dt 7:00 0.5 97 478 844 4578 55 80 54 550 cars

Eample A particle has velocity v(t) = t 0t + 4 t. Without evaluating, write an integral that represents the following quantities: a) Displacement over [0,6] 0 6 t 0t 4t dt b) Total distance traveled over [,5] 5 t 0t 4t dt

The Integral of Velocity Assume an object is in linear motion s(t) with velocity v(t). Since v(t) = s'(t): Displacement during, t t Distance traveled during t, t v t dt t t v t dt t t

White Board Challenge A factory produces bicycles at a rate of: p t 95 t t bicycles per week. How many bicycles were produced from the beginning of week to the end of week? Week Week Week Week 4 0 4 p t dt Bicycles

The Definite Integral as a Function of Let f be a continuous function on [a, b] and varies between a and b. If varies, the following is a function of denoted by g(): Notice that a is a real number. g a f t dt Signed Area = g() a b Notice, g() satisfies the initial condition g(a) = 0.

Eample Use the function F() to answer the questions below: F t dt a) Find a formula for the function. F b) Evaluate F(4). t dt 4 t 4 4 4 4 4 4 4 4 F 4 4 6.75 4 4 4 c) Find the derivative of F(). F ' 4 4 4 0 Notice that this is the same as the integral when t =.

Evaluate: Eample d d t dt In the previous eample, in order to find the derivative we had to find the integral: t dt Unfortunately, like many integrals, we can not find an antiderivative for this function. It should be clear there is an inverse relationship between the derivative and the integral. Thus, the derivative of the integral function is simply the original function.

The Second Fundamental Theorem of Calculus Assume that f() is continuous on an open interval I containing a. Then the area function: d d A a f t dt a is an antiderivative of f() on I; that is, A'() = f(). Equivalently, f t dt f

Eample (Continued) Evaluate: d d t dt f t t Since f() d d t dt

Evaluate: Eample d d sin t dt Notice the upper limit of the integral is a function of rather than itself. We can not apply the nd FTC. But we can find an antiderivative of the integral: sin t dt cost cos cos d cos cos Find the derivative of the result: cos cos d sin 0 sin

Eample 4 Evaluate: d d cos t dt Notice we can not find an antiderivative of the integral AND the upper limit of the integral is a function of rather than itself. How do we handle this? Can we apply the nd FTC?

The Upper Limit of the Integral is a Function of Use the First Fundamental Theorem of Calculus to evaluate the integral: g f t dt a Find the derivative of the result: d F g F a d F a F g d d Chain Rule d F a F g F ' g g ' 0 Constant d ' f g g

Composite Functions and The Second Fundamental Theorem of Calculus When the upper limit of the integral is a function of rather than itself: A g f t dt a We can use the Second Fundamental Theorem of Calculus together with the Chain Rule to differentiate the integral: d d a g ' f t dt f g g

Eample 4 (continued) Evaluate: d d cos t dt f t cos Since,, and t g g' d d cos t dt cos cos f(g()) g'()

White Board Challenge Find the derivative of the function: F sec t dt sec

White Board Challenge If h(t) is the rate of change of the height of a conical pile of sand in feet/hour, what does the following integral represent? Be specific. 6 h t dt The change in height of the pile of sand from hour to hour 6.

006 AB Free Response 4 Form B

99 AB Free Response Let f be the function that is defined for all real numbers and that has the following properties. i. f ''() = 4 8 ii. f '() = 6 iii. f () = 0 a) Find each such that the line tangent to the graph of f at (,f()) is horizontal. b) Write the epression for f(). c) Find the average value of f on the interval.

008 AB Free Response 5 Form B

0 AB Free Response