Fundamental Theorem of Calculus

R I E M A N N S U M S T R A P E Z O I D A L R U L E

D E F I N I T E I N T E G R A L S A V E R A G E M E A N V A L U E M E A N V A L U E T H E O R E M

I N T E R M E D I A T E V A L U E T H E O R E M F U N D A M E N T A L T H E O R E M O F C A L C U L U S

Fundamental Theorem of Calculus

Portfolio by Adeleine Tran AP Calculus BC | 2010

1

Overview

Fundamental Theorem of Calculus is not

only an important concept to grasp for

higher level math, it also provides useful

applications that can apply to real life

situation. Imagine if you want to find how

far you travel over a period of time with a

certain velocity, what would you do?

Before we get further into the Fundamental

Theorem of Calculus itself, we need to have

some basic background understanding.

First of all, we know that:

Velocity x Time = Distance

In the graph, the product of the x- (velocity)

and y-values (time) gives us the distance or

the displacement of the area traveled.

Thus, area represents distance moved:

(positive when v > 0, negative when v < 0).

tim

e

velocity

2

Riemann Sums

Riemann Sum is a method for

approximating the total area underneath a

curve on a graph, also known as an

integral. The sum of all areas represents

the total accumulated distance from time a

to b.

To find the total sum, you need to add up all

of the areas underneath the curves.

A Riemann Sum of f over [a, b] is the sum of

the areas of the rectangles formed between

the curves and the x-axis:

Rectangular Approximation Method (RAM)

is an example of Riemann Sums, which is

constructed to determine the total area of the

rectangles formed underneath the curves.

∑=

∆n

ii xxf

1)(

height base

3

Riemann Sums

Three ways of approximating the areas of the

rectangles are:

Midpoint Rectangular Approximation Method (MRAM)

Left-hand Rectangular Approximation Method (LRAM)

Right-hand Rectangular Approximation Method

(RRAM)

The name suggests the determining heights that

we need to use to approximate the area of the

rectangles. In other words, MRAM uses the mid-

point of the rectangle as the height, while LRAM

uses the top-left corner, and RRAM uses the top-

right corner.

Generally, MRAM is the most accurate of the

three. LRAM and RRAM might overestimate or

underestimate the approximation.

What LRAM, RRAM, and MRAM look like graphically:

LRAM

MRAM

RRAM

4

Riemann Sums

Compute the LRAM, RRAM, and MRAM of

the function

over [0, 2] with 4 subintervals.

EXAMPLE 1

22 xxy −=

SOLUTION

Computing LRAM:

The base is 0.5 because there are 4 subintervals

(2/4 = ½). The height is the y-value of the

function at the particular x-value.

Add the areas (base * height) together:

25.1)75.0)(5.0()1)(5.0()75.0)(5.0( =++=LRAM

5

Riemann Sums

Computing RRAM:

RRAM is the same as LRAM in this particular

case.

SOLUTION (cont.) Computing MRAM:

The height of the rectangle is the y-value of the

mid-point of each subinterval.

25.1)75.0)(5.0()1)(5.0()75.0)(5.0( =++=RRAM

375.1)4375.0)(5.0()9375.0)(5.0()9375.0)(5.0()4375.0)(5.0(

=+++=MRAM

6

Riemann Sums

If no function was given, another way to calculate

the sums is using a table.

Compute the (a) LRAM and (b) RRAM estimates

using 12 subintervals of length 5 to find how far

upstream the bottle travels.

LRAM:

RRAM:

EXAMPLE 2

Time (min)

Velocity (m/sec)

Time (min)

Velocity (m/sec)

0 1 35 1.2

5 1.2 40 1.0

10 1.7 45 1.8

15 2.0 50 1.5

20 1.8 55 1.2

25 1.6 60 0

30 1.4

SOLUTION

mssm 52206087

)2.15.18.10.12.14.16.18.127.12.11(5

=/×/

=

+++++++++++

mssm 49206082

)02.15.18.10.12.14.16.18.127.12.1(5

=/×/

=

+++++++++++

7

Trapezoidal Rule

A more efficient method in approximating

integrals than RAMs is by calculating the area of

trapezoids in order to find the area underneath

the curve of a function. This is known as the

Trapezoidal Rule.

where LRAM and RRAM are the Riemann sums

using the left and right endpoints, respectively.

Simply put, we find the area of trapezoids instead

of rectangles to get better approximation of the

areas under the curve.

The Trapezoidal Rule is also represented by:

where [a, b] is partitioned into n subintervals of

equal length and h = (b – a)/n

The area formula for trapezoid is

where h is the length of the subinterval. Thus,

when adding all of the trapezoids, except for the

first and last bases, the middle bases (which are

the y-values of the curve) repeat. Therefore, when

adding the bases together, you must account the

fact that there are 2 middle bases in between the

top and bottom bases of the trapezoid.

2nn RRAMLRAMT +

=

)2...22(2 1210 nn yyyyyhT +++++= −

))((21

21 bbh +

8

Riemann Sums and Integrals

• As n number of rectangles reaches infinity, the

Riemann Sums reaches its limit such that,

• As the partition gets finer, the Δx (the partition)

essentially tends to zero and becomes a differential

dx. The change in x-values has also became so

small that we could consider all x-values as

continuous in the interval of a to b. Because we are

summing all products of the areas, we can

abandon the k and n and set the limit as the

function goes from a to b.

• The expression is called Definite Integral.

∫∑ =∆=

∞→

b

a

n

kkn

dxxfxcf )()(lim1

As you can see from the illustrations, increasing

the number of equal-sized intervals the sum of the

areas under the curve can give better

approximation of the total area. Finer partitions of

the interval [a, b] create more rectangles with

shorter bases and increasing accuracy.

9

Definite Integrals

There are two ways to find the area under the curve: finding the areas of the geometric shapes, or

finding by taking the anti-derivative of the function.

Geometric shapes include triangles, rectangles, circles, trapezoids, and other shapes formed by the

enclosed area between the curve and the x-axis.

Anti-derivative , also known as indefinite integrals, of a function f (x) is a function F whose

derivative equals to f (x). We will further look at this concept later on.

Evaluate the integral:

Since the function f (x) represents the ¼ of a circle, we can find the area simply by using the area

formula of a circle where the radius = 4 (square root of 16) and divides the area by 4:

Rather than taking the painstaking anti-derivative of the derivative, finding the area using

geometric shape is easier in this case and still yields an accurate answer.

EXAMPLE ∫−

−0

4

216 x

πππ 4)4(41

41 22 === rA

SOLUTION

10

Definite Integrals

Area Under a Curve (as a Definite Integral)

If y = f (x) is nonnegative and integrable over a closed interval [a, b], then the area under the curve

y = f (x) from a to b is the integral of f from a to b,

And when the function is nonpositive, the Riemann sums for f over the interval [a, b] are negatives of

rectangle areas. Therefore,

If an integrable function y = f (x) has both positive and negative values, then the Riemann sums add

the positive and negative areas. Sometimes definite integral is called net area of the region because the

value of the integral is resulting area after the cancellation:

DEFNITION

∫=b

a

dxxfA )(

∫ ≤−=b

a

xfwhendxxfA 0)()(

∫ −−−=b

a

axisxthebelowareaaxisxtheaboveareadxxf )()()(

11

Basic Properties of Integrals

These properties of integrals follow from the definition of integrals as limits of Riemann sums.

Understanding these properties will further help us understand the proof of the FTC.

Zero:

Order of Integration:

Additivity:

Constant Multiple:

Sum and Difference:

( ) ( ) ( )f x f fb c b

a a c

dx x dx x dx= +∫ ∫ ∫3

( )f x 0c

c

dx =∫1

( ) ( )f x fb a

a b

dx x dx= −∫ ∫2

( ) ( )( ) ( ) ( )f x g f gb b b

a a a

x dx x dx x dx+ = +∫ ∫ ∫5

( ) ( )f x fb b

a a

r dx r x dx=∫ ∫4

12

Average (Mean) Value

To find the most area between the curve and the axis, we could use Riemann Sums and

rectangles to approximate the areas. However, using left-hand or right-hand corner of the

rectangle could either under or overestimate the actual area. This suggests that somewhere in

between, there is a “just right” height of the rectangle that would yield the most accurate

approximation of the area. The “just right” height is the average value of the function.

When multiplying the average value of the function (as the height) with the interval of the

function (as the base), the product (the area of the rectangle) is equal to the net area between

f and the x-axis.


If f is integrable on [a, b], its average (mean) value on [a, b] is

DEFNITION

∫−=

b

a

dxxfab

fav )(1)(

13


Find the average value of

the function on the interval. At what

point(s) in the interval does the function

assume its average value?

First, find the anti-

derivative, which equals to:

Apply the definition of average (mean)

value:

To find the point where the function

assumes its average value, set the original

function equals to the average value because

it represents the average height which is its

y-value. The function assumes its average

value at point c, where:

Since the limit is between 0 and 1, the

function assumes its average value at

EXAMPLE

SOLUTION

]1,0[,13)( 2 −−= xxf

xxF −−= 3

2)0()1(11

)13(01

1

)(1)(

1

0

2

−=−×=

−−−

=

−=

∫

∫

FF

dxx

dxxfab

favb

a

31

31

13213

2

2

2

±=

=

−=−

−=−−

c

c

cc

31

=c

14

Mean Value Theorem

If f is continuous on [a, b], then at some

point c in [a, b],

such that,

In other words, there is at least one point c

at which the derivative (slope) of the curve

is equal (parallel) to the average value of

the curve.

Meaning, there exists some c in the interval

(a, b) such that the secant joining the

endpoints of the interval [a, b] is parallel to

the tangent at c.

∫−=

b

a

dxxfab

cf )(1)(

abafbfcf

−−

=)()()('

15

Intermediate Value Theorem

Let f (x) be a continuous function on the

interval [a, b]. If d is in between the range

[f (a), f (b)], then there is a c in between the

domain [a, b] such that f (c) = d.

Simply put, Intermediate Value Theorem

states that if a particle moves in between the

interval [a, b] in a continuous function, it

will pass through every value in between

that is mapped by the function.

For example, if x travels from 30 to 70, it

must pass through 31, 32, 33, and so on

until it reaches 70.

16

Fundamental Theorem of Calculus Part I

The Fundamental Theorem of Calculus, Part 1

If f is continuous on [a, b], then the function

has a derivative at every point x in [a, b], and

The first part of the Fundamental Theorem of Calculus says that every continuous

function f is the derivative of some other functions, and every continuous function has an

anti-derivative. Simply put, it guarantees the existence of anti-derivative of continuous

functions.

Part 1 also specifies the relationship of indefinite integration and differentiation: the

processes of integration and differentiation are inverses of one another.

DEFNITION

∫=x

a

dttfxF )()(

)()()(' xfdttfdxdxF

x

a∫ ==

17


Suppose that f is continuous on [a, b].

Let

and that

Therefore, F(x + h) – F(x) is

PROOF

∫=x

a

dttfxF )()( ∫+

=+hx

a

dttfhxF )()(

∫∫∫++

=−=−+hx

x

x

a

hx

a

dttfdttfdttfxFhxF )()()()()(

Additivity Property of Integrals

18


Let M be the maximum and m be the minimum of f on [a, b] as shown:

Because M is the max and m is the min, any f (x) in the interval [x, x + h]

must be in between the lower and upper limits:

Thus, when finding the areas of the rectangle with base h, the rectangle with the height M produces the

largest approximation while height m produces the smallest area. The area under the curve in the interval

[x, x + h] is hence in between mh and Mh. If we divided by h, we get:

Substituting in F(x + h) – F(x):

Next, using limits to define the derivative of the areas:

As h 0, x + h = x + (0) = x. Therefore, while the derivative

of M is f (x), the derivative of m is also f (x + h) = f (x).

Mtfm ≤≤ )(

Mdttfh

mhMdttfhmh

hx

x

hx

x

≤≤=/≤≤// ∫∫

++

)(1))((1

MxFhxFh

mhx

x

≤−+≤ ∫+

)]()([1

MxFhxFh

mh

hx

xhh 000

lim)]()([1limlim→

+

→→≤−+≤ ∫

19


Since , we can apply the Sandwich Theorem which suggests that if

Therefore, if then

By the definition of derivative, the derivative of function F (x) with respect to the variable x is

By substitution, the derivative of the function is:

This completes the argument which justifies the relationship

)(limlim00

xfMmhh

==→→

FxfthenFxhxgandxhxfxgaxaxax

===≤≤→→→

)(lim,)(lim)(lim)()()(

),(limlim00

xfMmhh

==→→ ∫

+

=−+hx

x

xfxFhxFh

)()]()([1

hxFhxFxF

h

)()(lim)('0

−+=

→

∫=x

a

dttfxF )()(

)()()(lim)()('0

xfh

xfhxfdttfdxdxF

h

x

a

=−+

==→∫

)()()(' xfdttfdxdxF

x

a∫ ==

20


Now that we have proven part 1 of the FTC true, let’s see an example of its application.

Find the derivative of

Applying the FTC #1 to the problem, we can say that the derivative of y is cos (2x).

However, we must not forget the Chain Rule. Therefore, the answer is:

Understanding Part 1 is especially useful later on when we look at graphs and determine

their relationships to one another by knowing that every function has an anti-derivative

and is a derivative of some other functions.

EXAMPLE

∫=x

dtty2

0

cos

SOLUTION

xxxdttdxdy

x

2cos2)2()2cos(cos' 012

0

=×== ∫

21

Fundamental Theorem of Calculus Part II

The Fundamental Theorem of Calculus, Part 2

If f is continuous at every point [a, b], and if F is any anti-derivative of f on [a, b], then

This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

The second part allows one to compute the definite integral by using any one of its infinitely

many anti-derivatives. The definite integral of any continuous function f can be calculated

without taking limits, adding Riemann Sums, and with little effort as long as an anti-

derivative can be found. This second part of the Fundamental Theorem significantly

simplifies the computation of definite integral, or the area under the curve.

DEFNITION

∫ −=b

a

aFbFdxxf )()()(

22


Given two graphs, F (x) and f (x), where F is the anti-derivative, and f is the

definite integral.

*not drawn

to accuracy*

The area of f (x) is the definite integral which equals to

We can use the Riemann Sums in order to approximate the area of function f. By using the

Mean Value Theorem, we can assume that in between a and b, there is a “just right” c value

that would yield the best height (average value) for the most accurate area of the rectangle.

In the function F (x), the slope of the secant line that passes through a and b is

Based on the Mean Value Theorem, the average value is therefore:

PROOF

∫b

a

xf )(

.)()(ab

aFbF−−

.)()()(ab

aFbFcf−−

=

23


We can use the average (mean) value as the height and the interval [a, b] as the base to

find the area under the curve:

By the definition of definite integral, the area under the curve is defined as

Therefore,

This proves the second part of the theorem that states

)()()(

)()()(

aFbFab

aFbFabA

−=−−

×−=

dxxfAb

a∫= )(

).()()(),()()( aFbFdxxfthenaFbFdxxfAbecauseb

a

b

a

−=−== ∫∫

∫ −=b

a

aFbFdxxf )()()(

24


f is the differentiable function

whose graph is shown in the figure. The position at

time t (sec) of a particle moving along a coordinate

axis is

meters. Use the graph to answer the questions.

Give reasons for your answers.

(a) What is the particle’s velocity at time t = 3?

– Recall from our background knowledge about

derivative, velocity is the first derivative of the position.

In other words, velocity is derivative of the function s.

Based on the first part of the FTC,

Therefore, the particle’s velocity at time t = 3 is

(b) Is the acceleration of the particle at time t = 3

positive or negative?

– The acceleration is the second derivative of the

position, or the first derivative of the velocity. In part (a),

the velocity is Thus, the acceleration is

which is the slope of the function f.

EXAMPLE 1

∫=t

dxxfs0

)( ∫ ==t

tfdxxfdxds

0

)()('

0)3()(' === ftfs

At t = 3, the particle’s velocity is 0 unit/ sec.

).(' tfs =

)('" tfs =

Looking at the graph, at time t = 3, the slope is positive.

SOLUTION 1 25


(c) What is the particle’s position at time t = 3?

– The position can be calculated by finding the

displacement or the distance traveled. The area under

the curve can be computed easily by the geometric

method because the graph perfectly forms a triangle.

Since the area is below the x-axis, we must keep in

mind that it will be a negative value.

(d) When does the particle pass through the origin?

– Based on the second part of the FTC, we can say that

Thus, the displacement is the area above the curve [3,

6] subtracted by the area under the curve [0, 3].

( )( ) .96321

21 unitsbhA −=

−==

Looking at the graph, we can see that the area below is

equal to the area above. This means that the particles

moves 9 units away from the origin, and then at t = 3,

it moves back toward the origin and reaches the origin

at time t = 6 where it continues to move away in

positive direction.

(e) Approximately when is the acceleration zero?

– The acceleration is zero when the second derivative

of the position or the first derivative of the velocity is

zero. Looking at the graph, when t = 7. )0()6()(

6

0

FFdttf −=∫

∫ =−=6

0

0)0()6()( FFdttf

0)('" == tfs

The acceleration is zero at time t = 7.

26

Thus, the particle passes through the origin at t = 6 because

Fundamental Theorem of Calculus II

(f) When is the particle moving toward the origin?

Away from the origin?

– Looking at the graph, we can see that the area is

negative from [0, 3] and positive from t = 3 and

beyond. According to answer in part (d), the

particle moves away from the origin at [0, 3] and

toward the origin again at [3, 6]. After t = 6, the

particle continues moving in a positive direction,

meaning that it is moving away from the origin in

the same direction it has reached the origin.

Hence, the particle moves

(g) On which side of the origin does the particle lie

at time t = 9?

– Based on the answers in part (f), at time t > 6,

the particle moves away from the origin in the

positive direction. Therefore, at time t = 9, the

particle should still lie on the positive side.

As you can see, understanding the Fundamental

Theorem of Calculus makes computation of the

displacement easier and more convenience.

Rather than taking time to tediously calculate the

Riemann Sums, we can look at the graph, apply

the Theorem to the context, and come up with

accurate answers.

Away at 0 < t < 3 Toward at 3 < t < 6 Away at t > 6

At time t = 9, the particle lies on the positive side of the origin.

27


The graph of a function f

consists of a semicircle and two line segments as

shown below. Use the graph to answer the

questions. Let

(a) Find g (1).

– According to the FTC #2,

(b) Find g (3).

–The function is negative over the interval [1, 3] so the

area is also negative, according the definition of definite

integral. Hence, according to FTC #2,

However, we do not know what the anti-derivative is

equal to. Since the function g (x) is a definite integral,

we can instead evaluate the integral by finding the area

under the curve geometrically:

(c) Find g (–1).

– Similar to the method used in part (b), we have to use

geometry to evaluate the function. Notice that since the

interval is flipped to [1, -1] instead of [-1, 1], the area is

going to be negative according to Order of Integration.

EXAMPLE 2

SOLUTION 2

∫=x

dttfxg1

)()(

∫ =−==1

1

0)1()1()()1( FFdttfg

)]1()3([)()3(3

1

FFdttfg −−=−= ∫

( ) 11221)3( −=

××−=−= ∆Areag

( ) ππ −=

××−=−=− Ο

2241)1( Areag

28


(d) Find all values of x on the open interval (-3, 4) at

which g has a relative maximum.

– Relative maximum occurs at the first derivative of a

function is equal to zero and the second derivative is

less than zero. According to the FTC #1, the first

derivative of the integral g (x) is the function f (x).

The second derivative of g (x), which is the first

derivative of f (x), is the slope of the function f.

(e) Write an equation for the line tangent to the graph

of g at x = -1.

– The graph of the function f represents the slope of

the integral g (x). Since y = f (-1) = 2, the slope of g (x)

and its tangent line is 2. The tangent line is

According to the answer from part (c), at x = -1, the

function g equals to –π. Since the line is tangent to the

graph of g at x = -1, it must also share the same y-

value, –π. Knowing both the x and the y values, we

can therefore find b in the equation:

Hence, the equation for the line tangent to function g

is

(f) Find the coordinate of each point of inflection of

the graph g on the open interval (-3, 4).

– Points of inflection occur second derivative of g is

equal to zero. Since the graph y = f (x) represent the

first derivative of g, then the derivative (or the slope)

of f (x) is the second derivative of g. Looking at the

graph,

Looking at the graph, the relative maximum occurs at x = 1 because g’ (1) = f (1) = 0 and g” (1) = f ‘ (1) = negative.

bxy += 2

ππ −=−+−−=+−= 2)()1(22 yxb

π−+= 22xy

.2,10)(')(" =−=== xxatxfxg

29


(g) Find the range of g.

– The range of a function is the interval of the smallest

and greatest y-values of the function.

The smallest y-value of g is its minimum which occurs

at x = -3 and x = 3. Find g (-3) and g (3)

Since -2π < -1, the absolute minimum is at (-3, -2π).

The greatest y-value of g is its maximum which occurs

at x = 1. According to part (a), g (1) = 0.

Therefore, in a closed interval, the range of g is

Keep in mind that to find an area and to

integrate are different concepts. Finding the

area is computing the total displacement of the

graph as a whole by adding areas of each

section. To integrate is to find the net area of

the whole interval [a, b].

12121)()3(

2221)()3(

3

1

23

1

−=

××−==

−=

××−=−=−

∫

∫−

dttfg

dttfg ππ

]0,2[ π−

NOTE 30

Area v. Integrate

(a) Integrate the function over the interval and (b)

find the area of the region between the graph and

the x-axis:

An anti-derivative of the given function is

(a) Integrate to find the net area, 0 t0 3:

(b) Find the area by adding area above and below

the graph:

EXAMPLE

]3,0[,862 +−= xxy

SOLUTION

xxxxF 8331)( 23 +−=

∫ =−=−=+−3

0

2 606)0()3()86( FFdxxx

2

2

0

3

2

22

322)

32()

320(

)]2()3([)]0()2([

)86()86(

)()(

unit

FFFF

dxxxdxxx

AreaAreaAreaAreaArea

ba

belowaboveregion

=−−=

−−−=

+−−+−=

−+=

+=

∫ ∫

31

Summary

With a strong understanding of the Fundamental Theorem of Calculus, we are able to

figure out the relationship between different functions with little efforts. This knowledge

also transfers to multiple real-life applications involving relationships such as those

between acceleration, velocity, and position of moving objects. The Fundamental

Theorem of Calculus allows us to calculus the area under the curve and the displacement

of a function with optimal accuracy that does not require Riemann Sums nor using limits.

Not only so, the Fundamental Theorem of Calculus ties two different branches of

mathematics together: differential and integral, and with this insight, the Fundamental

Theorem of Calculus becomes a powerful tool for understanding how the universe tied

together.

32

Fundamental Theorem of Calculus

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