Fundamental Theorem of Calculus by Adeleine Tran

Fundamental Theorem of Calculus 1

RIEMANN SUMS TRAPEZOIDAL RULE DEFINITE INTEGRALS AVERAGE MEAN VALUE MEAN VALUE THEOREM INTERMEDIATE VALUE THEOREM FUNDAMENTAL THEOREM OF CALCULUS Portfolio by Adeleine Tran AP Calculus BC | 2010

Overview 2  Fundamental Theorem of Calculus is not

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

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

higher level math, it also provides useful

the displacement of the area traveled.

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?

time

only an important concept to grasp for

velocity  Thus, area represents distance moved:

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

Riemann Sums 3 

Riemann Sum is a method for



approximating the total area underneath a curve on a graph, also known as an

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

integral. The sum of all areas represents

the areas of the rectangles formed between

the total accumulated distance from time a

the curves and the x-axis:

to b.

∑ f ( x )∆x i =1

height 

base

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.

Riemann Sums 4 

Three ways of approximating the areas of the



What LRAM, RRAM, and MRAM look like graphically:

rectangles are: 

Midpoint Rectangular Approximation Method (MRAM)



Left-hand Rectangular Approximation Method (LRAM)



Right-hand Rectangular Approximation Method

 LRAM

(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-

RRAM 

point of the rectangle as the height, while LRAM uses the top-left corner, and RRAM uses the topright corner. 

Generally, MRAM is the most accurate of the three. LRAM and RRAM might overestimate or underestimate the approximation.

 MRAM

Riemann Sums 5

SOLUTION

EXAMPLE 1 

Compute the LRAM, RRAM, and MRAM of



Computing LRAM:



The base is 0.5 because there are 4 subintervals

the function

y = 2x − x2 over [0, 2] with 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:

LRAM = (0.5)(0.75) + (0.5)(1) + (0.5)(0.75) = 1.25

Riemann Sums 6

SOLUTION (cont.) 



Computing MRAM:



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

Computing RRAM:

mid-point of each subinterval.

RRAM = (0.5)(0.75) + (0.5)(1) + (0.5)(0.75) = 1.25 

RRAM is the same as LRAM in this particular case.

MRAM = (0.5)(0.4375) + (0.5)(0.9375) + (0.5)(0.9375) + (0.5)(0.4375) = 1.375

Riemann Sums 7 

If no function was given, another way to calculate the sums is using a table.

SOLUTION 

EXAMPLE 2 

5(1 + 1.2 + 1.7 + 2 + 1.8 + 1.6 + 1.4 + 1.2 + 1.0 + 1.8 + 1.5 + 1.2) m = 87 × 60 s/ = 5220m s/

Compute the (a) LRAM and (b) RRAM estimates using 12 subintervals of length 5 to find how far upstream the bottle travels. Time (min)

Velocity (m/sec)

Time (min)

Velocity (m/sec)

1.2

1.0

1.7

1.8

2.0

1.5

1.8

1.2

1.6

1.4

LRAM:



RRAM:

5(1.2 + 1.7 + 2 + 1.8 + 1.6 + 1.4 + 1.2 + 1.0 + 1.8 + 1.5 + 1.2 + 0) m = 82 × 60 s/ = 4920m s/

Trapezoidal Rule 8 

A more efficient method in approximating



integrals than RAMs is by calculating the area of

The Trapezoidal Rule is also represented by:

h T = ( y0 + 2 y1 + 2 y2 + ... + 2 yn −1 + yn ) 2

trapezoids in order to find the area underneath the curve of a function. This is known as the

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

Trapezoidal Rule.

equal length and h = (b – a)/n

LRAM n + RRAM n T= 2



The area formula for trapezoid is

1 (h)(b1 + b2 ) 2 where h is the length of the subinterval. Thus,

where LRAM and RRAM are the Riemann sums using the left and right endpoints, respectively. 

Simply put, we find the area of trapezoids instead

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

of rectangles to get better approximation of the

fact that there are 2 middle bases in between the

areas under the curve.

top and bottom bases of the trapezoid.

Riemann Sums and Integrals 9 

As you can see from the illustrations, increasing

•

Riemann Sums reaches its limit such that,

the number of equal-sized intervals the sum of the areas under the curve can give better

k =1

lim ∑ f (ck ) ∆x = ∫ f ( x) dx

approximation of the total area. Finer partitions of

n →∞

the interval [a, b] create more rectangles with shorter bases and increasing accuracy.

As n number of rectangles reaches infinity, the

•

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.

Definite Integrals 10 

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.

EXAMPLE

Evaluate the integral:

∫

16 − x 2

−4

SOLUTION 

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:

A= 

1 1 π r 2 = π (4) 2 = 4π 4 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.

Definite Integrals 11

DEFNITION 

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, b

∫

f ( x ) dx



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

A = − ∫ f ( x ) dx

when f ( x ) ≤ 0



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: b

∫ f ( x) dx = (area above the x − axis) − (area below the x − axis) a

Basic Properties of Integrals 12 

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:

∫ f ( x ) dx = 0 c

Order of Integration:

Additivity:

3 Constant Multiple:

4 Sum and Difference:

∫ f ( x ) dx = − ∫ f ( x ) dx b

f ( x ) dx ∫ f ( x ) dx + ∫ f ( x ) dx ∫= b

∫ r f ( x ) dx = r ∫ f ( x ) dx b

∫ ( f ( x ) + g ( x ) ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx

Average (Mean) Value 13  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.

DEFNITION

Average (Mean) Value

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

1 av ( f ) = b−a

∫ a

f ( x ) dx

Average (Mean) Value 14

EXAMPLE

Find the average value of

 To find the point where the function

the function on the interval. At what

assumes its average value, set the original

point(s) in the interval does the function

function equals to the average value because

assume its average value?

it represents the average height which is its

f ( x) = −3x 2 − 1, [0,1] SOLUTION

y-value. The function assumes its average

First, find the anti-

derivative, which equals to: F = − x − x 3

 Apply the definition of average (mean)

value:

1 av ( f ) = b−a

∫ f ( x) dx a

1 2 x = − − 1) dx ( 3 1 − 0 ∫0 1 = × F (1) − F (0) = −2 1

value at point c, where: − 3c 2 − 1 = −2

− 3c 2 = −1 c2 =

1 3

1 3  Since the limit is between 0 and 1, the c=±

function assumes its average value at c =

1 3

Mean Value Theorem 15  If f is continuous on [a, b], then at some

point c in [a, b],

1 f (c ) = f ( x) dx b − a ∫a

such that,

f ' (c ) =

f (b) − f ( a ) b−a

 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.

Intermediate Value Theorem 16  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.

Fundamental Theorem of Calculus Part I 17

DEFNITION

The Fundamental Theorem of Calculus, Part 1

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

F ( x) = ∫ f (t ) dt a

d F ' ( x) = ∫ f (t ) dt = f ( x) has a derivative at every point x in [a, b], and dx a  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.

Fundamental Theorem of Calculus Part I 18

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

x+h

F ( x) = ∫ f (t ) dt

and that

F ( x + h) =

∫ f (t ) dt a

 Therefore, F(x + h) – F(x) is

F ( x + h) − F ( x ) =

x+h

∫ f (t ) dt − ∫ f (t ) dt = ∫ f (t ) dt

Additivity Property of Integrals

Fundamental Theorem of Calculus Part I 19 

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: m ≤ f (t ) ≤ M



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:

1 (mh/ ≤ h/

x+h



1 / ( ) ) ≤ = ≤ f t dt M h m f (t ) dt ≤ M ∫x h ∫x x+h 1 Substituting in F(x + h) – F(x): m ≤ [ F ( x + h) − F ( x)] ≤ M h ∫x x+h 1 lim m ≤ lim ∫ [ F ( x + h) − F ( x)] ≤ lim M h →0 h h →0 Next, using limits to define the derivative of the areas: h→0 x



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).

Fundamental Theorem of Calculus Part I 20 

m = lim M = f ( x) , we can apply the Sandwich Theorem which suggests that if Since lim h →0 h →0

g ( x) ≤ f ( x) ≤ h( x) and lim g ( x) = lim h( x) = F , then lim f ( x) = F x→a

 

Therefore, if

lim m = lim M = f ( x), h →0

h →0

then

x→a

1 h

x→a

x+h

∫ [ F ( x + h) − F ( x)] = f ( x) x

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

F ' ( x) = lim h →0

F ( x + h) − F ( x ) h x



By substitution, the derivative of the function

F ( x) = ∫ f (t ) dt is: a

d f ( x + h) − f ( x ) ( ) lim f t dt F ' ( x) = = = f ( x) ∫ 0 h → dx a h x



This completes the argument which justifies the relationship

d F ' ( x) = f (t ) dt = f ( x) dx ∫a

Fundamental Theorem of Calculus Part I 21  Now that we have proven part 1 of the FTC true, let’s see an example of its application.

EXAMPLE

Find the derivative of

y = ∫ cos t dt 0

SOLUTION  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: 2x

d y '= cos t dt = cos(2 x1 ) × (2 x 0 ) = 2 cos 2 x ∫ dx 0  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.

Fundamental Theorem of Calculus Part II 22

DEFNITION

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 b

∫ f ( x) dx = F (b) − F (a) a

 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 antiderivative can be found. This second part of the Fundamental Theorem significantly simplifies the computation of definite integral, or the area under the curve.

Fundamental Theorem of Calculus Part II 23

PROOF

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

definite integral. *not drawn to accuracy* b

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

∫ f ( x) a

 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 F (b) − F (a ) . b−a  Based on the Mean Value Theorem, the average value is therefore:

f (c ) =

F (b) − F (a ) . b−a

Fundamental Theorem of Calculus Part II 24  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:

F (b) − F (a ) (b − a ) = F (b) − F (a )

A = (b − a ) ×

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

A = ∫ f ( x) dx a

because A = ∫ f ( x) dx = F (b) − F (a ), then ∫ f ( x) dx = F (b) − F (a ).

 This proves the second part of the theorem that states b

∫ f ( x) dx = F (b) − F (a) a

Fundamental Theorem of Calculus Part II 25

EXAMPLE 1

SOLUTION 1

f is the differentiable function

whose graph is shown in the figure. The position at



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

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

– Recall from our background knowledge about

axis is

derivative, velocity is the first derivative of the position. t

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

s = ∫ f ( x) dx

d f ( x) dx = f (t ) Based on the first part of the FTC, s ' = dx ∫0

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

meters. Use the graph to answer the questions.

s ' = f (t ) = f (3) = 0

Give reasons for your answers. At t = 3, the particle’s velocity is 0 unit/ sec. 

(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

s ' = f (t ). Thus, the acceleration is

s " = f ' (t ) which is the slope of the function f. Looking at the graph, at time t = 3, the slope is positive.

Fundamental Theorem of Calculus Part II 26 



– The position can be calculated by finding the

equal to the area above. This means that the particles

displacement or the distance traveled. The area under

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

the curve can be computed easily by the geometric

it moves back toward the origin and reaches the origin

method because the graph perfectly forms a triangle.

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

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

positive direction.

mind that it will be a negative value.

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

∫ f (t ) dt = F (6) − F (0) = 0

1 1 A = bh = − (3)(6 ) = −9 units. 2 2 

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

(d) When does the particle pass through the origin? – Based on the second part of the FTC, we can say that 6

∫

f (t ) dt = F (6) − F (0)

Thus, the displacement is the area above the curve [3, 6] subtracted by the area under the curve [0, 3].



(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,

s " = f ' (t ) = 0 when t = 7.

The acceleration is zero at time t = 7.

Fundamental Theorem of Calculus II 27 

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



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

Away from the origin?

at time t = 9?

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

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

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

the particle moves away from the origin in the

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

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

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

particle should still lie on the positive side.

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

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

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



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.

Fundamental Theorem of Calculus II 28

EXAMPLE 2

The graph of a function f



(b) Find g (3).

consists of a semicircle and two line segments as

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

shown below. Use the graph to answer the

area is also negative, according the definition of definite

questions. Let

integral. Hence, according to FTC #2, 3

g (3) = − ∫ f (t ) dt = −[ F (3) − F (1)]

g ( x) = ∫ f (t ) dt

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:

SOLUTION 2 

(a) Find g (1). – According to the FTC #2, 1

g (1) = ∫ f (t ) dt = F (1) − F (1) = 0 1



 1 g (3) = −( Area∆ ) = − × 2 × 1 = −1  2 (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.  1 g (−1) = −( AreaΟ ) = − × π × 2 2  = −π  4

Fundamental Theorem of Calculus II 29 

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



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

which g has a relative maximum.

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

– Relative maximum occurs at the first derivative of a

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

function is equal to zero and the second derivative is

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

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

can therefore find b in the equation:

b = −2 x + y = −2(−1) + (−π ) = 2 − π

derivative of the integral g (x) is the function f (x). The second derivative of g (x), which is the first

Hence, the equation for the line tangent to function g

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

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



y = 2x + 2 − π

(f) Find the coordinate of each point of inflection of the graph g on the open interval (-3, 4).

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

– Points of inflection occur second derivative of g is

of g at x = -1.

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

– The graph of the function f represents the slope of

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

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

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

and its tangent line is 2. The tangent line is

graph,

y = 2x + b

g " ( x) = f ' ( x) = 0 at x = −1, x = 2.

Fundamental Theorem of Calculus II 30 

NOTE

(g) Find the range of g. – The range of a function is the interval of the smallest



Keep in mind that to find an area and to

and greatest y-values of the function.

integrate are different concepts. Finding the

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

area is computing the total displacement of the

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

graph as a whole by adding areas of each

−3

 1 g (−3) = − ∫ f (t ) dt = − × π × 2 2  = −2π  2 1  1 g (3) = ∫ f (t ) dt = − ×1× 2  = −1  2 1 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

[−2π , 0]

section. To integrate is to find the net area of the whole interval [a, b].

Area v. Integrate 31

EXAMPLE 

(a) Integrate the function over the interval and (b) find the area of the region between the graph and the x-axis:

y = x 2 − 6 x + 8, [0, 3]

SOLUTION An anti-derivative of the given function is 1 F ( x) = x 3 − 3x 2 + 8 x 3  (a) Integrate to find the net area, 0 t0 3: 

∫ (x

− 6 x + 8) dx = F (3) − F (0) = 6 − 0 = 6



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

Arearegion = Areaabove + Areabelow = ( Areaa ) + (− Areab ) 2

= ∫ ( x − 6 x + 8) dx − ∫ ( x 2 − 6 x + 8) dx 2

= [ F (2) − F (0)] − [ F (3) − F (2)] =(

20 2 22 ) − (− ) = unit 2 3 3 3

Summary 32 ď&#x201A;&#x2014; 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.