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 '(x) = f(x) throughout this interval then

b

af x dx F b F a

Alternative forms: '

b

af x dx f b f a

b

aF b F a f x dx

Example 1Evaluate

First Find the indefinte integral

F(x): 2F x x dx 2 11

2 1 x C 31

3 x C

13

Now apply the FTC to find the definite

integral:

F b F a 1 0F F

3 31 13 31 0C C

13 C C Notice that it is

not necessary to include the “C”

with definite integrals

1 2

0x dx

Example 2Evaluate

First Find the indefinte integral

F(x): 3 2F x x dx 3 11

3 1 2x x C

414 2x x C

394

Now apply the FTC to find the definite

integral:

F b F a 2 1F F

4 41 14 42 2 2 1 2 1C C

748 C C Notice that it is

not necessary to include the “C”

with definite integrals

2 3

12x dx

2

2cos x dx

More Examples: New Notation4. Evaluate

2

2sin x

2 2sin sin 1 1 F(x) Bounds

2

5

103 xx x dx

5. Evaluate

5313 3

10x x

331 13 35 10 5 3 10 3 25

3 21 883

91

4 xx dx6. Evaluate

91 2 212 4

2x x 1 2 2 1 2 21 12 22 9 9 2 4 4

34.5

9 1 2

4x x dx

5 2

310x dx

If needed, rewrite.

Example 7Calculate the total area between the curve y = 1 – x2 and the

x-axis over the interval [0,2].The question considers all

area to be positive (not signed area), thus use the absolute value function:

21 x

2

2

2

1 , 1

1 , 1 1

1 , 1

x x

x x

x x

2

Use a integral and the piece-wise function to find the area:2 2

01 x dx 1 22 2

0 11 1x dx x dx

3 31 2

3 30 1

x xx x

33 3 301 2 13 3 3 31 0 2 1

Rewrite the equation as a

piece-wise function.

2 43 3

Example 8Assume F '(x) = f (x), f (x) = sin (x2), and F(2) = -5. Find

F (1).Use the First Fundamental

Theorem of Calculus: 2 2

1sin 2 1x dx F F

2 2

1sin 5 1x dx F

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.

0.495 5 1F

1 5.495F

Example 9The graph below is of the function f '(x).

If f (4) = 3, find f (12).Use the First Fundamental

Theorem of Calculus: 12

4' 12 4f x dx f f

212 4 12 3f

12 3 8f

White Board ChallengeIf, for all x, f '(x) = (x – 2)4(x – 1)3, it follows that the function f has:

a) a relative minimum at x = 1.b) a relative maximum at x = 1.c) both a relative minimum at x = 1 and a

relative maximum at x = 2.d) neither a relative maximum nor a relative

minimum.e) relative minima at x = 1 and at x = 2.

Multiple C

hoice

F b F ab af c

Let F be a function that satisfies the following hypotheses:1. F is continuous on the closed interval [a,b]2. F is differentiable on the open interval (a,b)Then there is a number c in (a,b) such that:

Mean Value TheoremLet f be a function that satisfies the following hypotheses:1. f is continuous on the closed interval [a,b]2. f is differentiable on the open interval (a,b)Then there is a number c in (a,b) such that:

' f b f ab af c

b

af x dx

b af c

b

ab a f c f x dx

Redefine the Conditions

Rewrite with integral notation.

Solve for the integral.

Mean Value Theorem for Integrals

If f(x) is continuous on [a,b], then there exists a value c on the interval [a,b] such that:

b

af x dx b a f c

Average Value of a FunctionThe average value of an integrable function f(x) on [a,b] is the quantity:

1 b

af x dx

b a

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 ab a

Approximates the derivative of a function.

Example 1Find the average value of f (x) = sin x on [0,π].

Use the Formula:

0

1 sin0

x dx

0

1 cos x

1 cos cos0

1 1 1

2

Example 2The height of a jump of a bushbaby is modeled by h (t)

= v0t – ½gt2. If g = 980 cm/s2 and the initial velocity is v

0=600

cm/s, find the average speed during the jump.

Use the Average Value Formula: 1 b

af x dx

b a

210 2h t v t gt

212600 980h t t t

2600 490h t t t

We are trying to find the average value of SPEED (absolute value of velocity).

So we need to find the velocity function.

Velocity is the derivative of Position.

'v t h tThus, the speed function is:

600 980t

Now find when the jump begins and ends (a and b).

20 600 490t t 60490,t

Evaluate the integral:6049

60 049

1 600 9800

t dt

Example 2 (Continued)

Rewrite the equation as a piece-wise function: 600 980t

3049

30 6049 49

600 980 , 0600 980 ,

t tt t

300 /cm s

Use a integral and the piece-wise function to find the average value:6049

60 049

1 600 980t dt 30 6049 49

3049

4960 0

600 980 600 980t dt t dt

30 6049 49

3049

2 24960 0

600 490 600 490t t t t 2 2 2249 30 30 60 60 30 30

60 49 49 49 49 49 49600 490 600 0 490 0 600 490 600 490 49 1800060 49

The height of a jump of a bushbaby is modeled by h (t) = v

0t – ½gt2. If g = 980 cm/s2 and the initial velocity is v

0=600

cm/s, find the average speed during the jump.

White Board Challenge

Evaluate

2

0

coslimh

hh

1

Net Change of a Quantity over a Specified Interval

Consider the following problems:1. Water flows into an empty bucket at a rate of

1.5 liters/second. How much water is in the bucket after 4 seconds?

2. Suppose the flow rate varies with time and can be represented as r(t). How much water is in the bucket after 4 seconds?

Quantity of water flow rate time elapsed 1.5 4 6 liters

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 1.5 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?

4

0' 4 0s t dt s s

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:

4

04 0r t dt s

4

04s r t dt

Signed Area under the graphWater in the bucket at 4 s

IMPORTANT:If the bucket did not

start empty, the integral would

represent the net change of water.

Net Change as the Integral of a Rate

The net change in s(t) over an interval [t1,t

2]

is given by the integral:

2

12 1'

t

ts t dt s t s t

Integral of the rate of change

Net change from t1 to t

2

Rate at which s(t) is changing

Amount of the quantity at t

1

Example 1If 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

ab t dt

The increase in the number of bacteria from hour a to hour c.

Example 2The 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 2-hour period.

t 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00

q(t) 1044 1297 1478 1844 1451 1378 1155 802 542

Since there is no function, we can not use the First Fundamental Theorem of Calculus. Instead approximate the area under the curve with any

Riemann Sum (I will use right-endpoints with 0.25 hour lengths):

9:00

7:00q t dt

0.25 1297 1478 1844 1451 1378 1155 802 542

2550 cars

Example 3A particle has velocity v(t) = t3 – 10t2 + 24 t. Without evaluating, write an integral that represents the following quantities:a) Displacement over [0,6]

b) Total distance traveled over [3,5]

6 3 2

010 24t t t dt

5 3 2

310 24t t t dt

The Integral of VelocityAssume an object is in linear motion s(t) with velocity v(t). Since v(t) = s'(t):

2

1

2

1

1 2

1 2

Displacement during ,

Distance traveled during ,

t

t

t

t

t t v t dt

t t v t dt

White Board ChallengeA factory produces bicycles at a rate of:

bicycles per week. How many bicycles were produced from the beginning of week 2 to the end of week 3?

295 3p t t t

3

1212p t dt Bicycles

Week 1 Week 2 Week 3 Week 40 1 2 3 4

a x b

Signed Area = g(x)

The Definite Integral as a Function of x

Let f be a continuous function on [a, b] and x varies between a and b. If x varies, the following is a function of x denoted by g(x):

x

ag x f t dt

Notice, g(x) satisfies the initial condition g(a) = 0.

Notice that a is a real number.

Example 1Use the function F(x) to answer the questions below:

3

1

xF x t dt

a) Find a formula for the function.

b) Evaluate F(4).

c) Find the derivative of F(x).

3

1

xF x t dt 41

4 1

xt 4 41 1

4 4 1x 41 14 4x

4F 41 14 44 63.75

'F x 4 114 4 0x 3x Notice that this is the same

as the integral when t = x.

Example 2Evaluate:

3

21

xd t dtdx

In the previous example, in order to find the derivative we

had to find the integral:

3

21

xt 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(x) is continuous on an open interval I containing a. Then the area function:

x

a

d f t dt f xdx

x

aA x f t dt

is an antiderivative of f(x) on I; that is, A'(x) = f(x). Equivalently,

Example 2 (Continued)Evaluate:

3

21

xd t dtdx

Since 31f t t

3 3

21 1

xd t dt xdx

f(x)

Example 3Evaluate:

2

2sin

xd t dtdx

Notice the upper limit of the integral is a function of x rather than x itself. We can not apply the 2nd FTC. But

we can find an antiderivative of the integral:2

2sin

xt dt

2

2cos

xt

2cos cos 2x

2cos cos 2x Find the derivative of the result:

2cos cos 2d xdx

2sin 2 0x x 22 sinx x

Example 4Evaluate:

13

1cosxd t dt

dx

Notice we can not find an antiderivative of the integral AND the upper limit of the integral is a function of x

rather than x itself.

How do we handle this? Can we apply the 2nd FTC?

The Upper Limit of the Integral is a Function of x

Use the First Fundamental Theorem of Calculus to evaluate the integral:

d F g x F adx

g x

af t dt

Find the derivative of the result:

F g x F a

d dF g x F adx dx

' ' 0F g x g x

'f g x g x

Chain Rule Constant

Composite Functions and The Second Fundamental Theorem of Calculus

When the upper limit of the integral is a function of x rather than x itself:

'g x

a

d f t dt f g x g xdx

g x

aA x f t dt

We can use the Second Fundamental Theorem of Calculus together with the Chain Rule to differentiate the integral:

Example 4 (continued)Evaluate:

13

1cosxd t dt

dx

Since , , and 3cosf t t

1

23 3 1 1

1cos cosx

x x

d t dtdx

f(g(x))

1xg x 2

1'x

g x

g'(x)

3 1

2

cos x

x

White Board ChallengeFind the derivative of the function:

2

1 secx

F x t dt

2 1 sec 2x

White Board ChallengeIf 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

2h t dt

The change in height of the pile of sand from hour 2 to hour 6.

2006 AB Free Response 4 Form B

1991 AB Free Response 1Let f be the function that is defined for all real numbers x and that has the following properties.i. f ''(x) = 24x – 18ii. f '(1) = –6iii. f (2) = 0

a) Find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal.

b) Write the expression for f(x).c) Find the average value of f on the interval

1≤x≤3.

2008 AB Free Response 5 Form B

2012 AB Free Response 1