The Definite Integral

Riemann Sums: In the previous set of notes we calculated the area under a curve by finding the sumof n rectangles. Each of these sums was an example of a more general sum called a Riemann Sum. We will now look more closely at this

subinterval will be called the rectangle. These rectangles may be above or below the xkth rectangle touches the curve The process for finding area between the xhave already learned. To find the area of the kth rectangle, we must find a way to represent its width and height

by )( kxf . Then the area of the kth rectangle is

all the rectangles under the curve is represented by the sum

“ PS ” stands for the sum of the partition.see it really is no different from the sums we have been using except that it is more general. Notable Differences from Previous Examples:

1. In previous examplesrepresented by∆

2. These rectangles may be above or below the xexamples.

3. The point where each rectangle intersects thewas in previous examples where we created rectangles that intersected the curve at an endpoint or midpoint of the rectangle.

So a Riemann Sum is simply a generalized version of the sums we have already worked with in previous examples but now we have a name to attach to these sums.

The Definite Integral

In the previous set of notes we calculated the area under a curve by finding the sum. Each of these sums was an example of a more general sum called a

Riemann Sum. We will now look more closely at this concept.

To create a Riemann Sum wwith a function

(fy =defined over a closed interval [a, b]. We will then divide this interval into subintervals, not necessarily of equal widths. This division of the interval into sucalled a [a, b].

subinterval will be called the kth subinterval and will be represented by the kth These rectangles may be above or below the x-axis. The point where the

kth rectangle touches the curve may vary from rectangle to rectangle.

he process for finding area between the x-axis and the curve is similar to what we To find the area of the kth rectangle, we must find a way to

height. The width will be represented by kx∆area of the kth rectangle is kkk xxfA ∆= )( and the

all the rectangles under the curve is represented by the sum ∑=

=n

kPS

1

the sum of the partition. This sum is a Riemann Sum. As you can see it really is no different from the sums we have been using except that it is more

ble Differences from Previous Examples: In previous examples the width of each rectangle was constant and was

x∆ but this is not important with a Riemann Sum.These rectangles may be above or below the x-axis unlike in previous

The point where each rectangle intersects the curve is also not important as it was in previous examples where we created rectangles that intersected the curve at an endpoint or midpoint of the rectangle.

So a Riemann Sum is simply a generalized version of the sums we have already revious examples but now we have a name to attach to these sums.

In the previous set of notes we calculated the area under a curve by finding the sum . Each of these sums was an example of a more general sum called a

To create a Riemann Sum we will begin with a function

)(x which is defined over a closed interval [a, b]. We will

ivide this interval into subintervals, not necessarily of equal widths. This division of the interval [a, b] into subintervals is called a Partition of

Each be represented by the kth

The point where the may vary from rectangle to rectangle.

axis and the curve is similar to what we To find the area of the kth rectangle, we must find a way to

and its height

and the total area of

∆ kk xxf )( where

This sum is a Riemann Sum. As you can see it really is no different from the sums we have been using except that it is more

the width of each rectangle was constant and was but this is not important with a Riemann Sum.

axis unlike in previous

curve is also not important as it was in previous examples where we created rectangles that intersected the

So a Riemann Sum is simply a generalized version of the sums we have already revious examples but now we have a name to attach to these sums.

So far in our discussion of Riemann Sums we have not mentioned limits. In order to find the exact area under a curve and not just an approximation, it was necessary to take the limit as the number of rectangles approached infinity. We will now do that with a Riemann Sum, but instead of letting the number of rectangles approach infinity, we will let the norm of the partition

The norm of the partition is the rectangle of greatest width. Consequently, if the norm of the partition is approaching zero, the width of also be approaching zero. This gives the following

This is the limit of a Riemann Sum. This special limit is extremely important in the study of calculus and is given its own name, the The Definite Integral: Unlike the indefinite integral we learned about when finding antiderivatives,definite integral is the limit of a Riemann Sum. Here is an explanation of the notation for the Definite Integral. Since the Definite Integral is the limit of a Riemann Sum, it may be used to find the area under a curve and in fact thefunction is defined to be a definite integral.

So far in our discussion of Riemann Sums we have not mentioned limits. In order to find the exact area under a curve and not just an approximation, it was necessary to

he number of rectangles approached infinity. We will now do that with a Riemann Sum, but instead of letting the number of rectangles approach infinity, we will let the norm of the partition P approach zero. →P

The norm of the partition is the rectangle of greatest width. Consequently, if the norm of the partition is approaching zero, the width of every other rectanglealso be approaching zero. This gives the following limit.

∫∑ =∆=→

b

a

n

kk

PdxxfxxfLim )()(

10

This is the limit of a Riemann Sum. This special limit is extremely important in the study of calculus and is given its own name, the Definite Integral.

The Definite Integral:

Unlike the indefinite integral we learned about when finding antiderivatives,definite integral is the limit of a Riemann Sum. Here is an explanation of the notation for the Definite Integral.

ince the Definite Integral is the limit of a Riemann Sum, it may be used to find the area under a curve and in fact the area under the graph of a non-negative continuous function is defined to be a definite integral.

So far in our discussion of Riemann Sums we have not mentioned limits. In order to find the exact area under a curve and not just an approximation, it was necessary to

he number of rectangles approached infinity. We will now do that with a Riemann Sum, but instead of letting the number of rectangles approach

0→

The norm of the partition is the rectangle of greatest width. Consequently, if the every other rectangle must

This is the limit of a Riemann Sum. This special limit is extremely important in the

Unlike the indefinite integral we learned about when finding antiderivatives, the definite integral is the limit of a Riemann Sum. Here is an explanation of the notation

ince the Definite Integral is the limit of a Riemann Sum, it may be used to find the negative continuous

EXAMPLE 1: Find the area under the curve of the function 2)( += xxf over the interval [0, 10] using the definite integral.

SOLUTION: This problem requires us to set up the integral ∫ +10

0

)2( dxx and

evaluate it using Riemann Sums.

∑∫=→

∆=n

kk

P

b

a

xxfLimdxxf1

0)()(

To do this is easier than you might think. In fact the procedure is almost exactly the same as in previous problems. Step 1: Determine the width of the kth rectangle by finding x∆ . Although a Riemann Sum does not require a constant width rectangle, we are not prohibited from doing it that way.

nnn

abx

10010 =−=−=∆

Step 2: Use the REP formula xkax ∆+= to determine the height )( xkafh ∆+= of the kth rectangle. Again, we are not required to use a specific point on each rectangle but it is perfectly acceptable to do so.

REP = n

kk

nxka

10100 =+=∆+ 2

1010)( +=

=∆+n

k

n

kfxkaf

Step 3: Find the area of the kth rectangle xkxafAk ∆−∆+= ))1(( .

nn

k

nn

kxxkafAk

20100102

10)(

2+=

+=∆∆+=

Step 4: Find the total area of all n rectangles ∑=

=n

kkT AA

1

.

Area = ∑∑∑∑∑∞

=

∞

=

∞

=

∞

=

∞

=

+=

+

=

+11

211

21

21

201002010020100

kkkkk nk

nnn

k

nn

k

= ( )nnn

nn

n

nn

n

507020

505020

)1(5020

2

)1(1002

+=++=++=+

+

Step 5: Find the limit of the sum as n approaches infinity.

Area = ( ) 7007050

7050

70 =+=

+=

+∞→∞→∞→ nn LimLimLim

nnn

Therefore 70)2(10

0

=+∫ dxx .

Properties of Definite Integrals: All continuous functions are integrable. That is, if a function )(xf is continuous over an interval [a, b] then its definite integral over [a, b] exists. The following properties of definite integrals apply to integrable functions. They may not all seem very important at this time, but they will be in the near future when we continue our discussion of area.

This property switches the bounds of integration.

Example: If ∫ =8

1

25)( dxxf , then∫ −=1

8

25)( dxxf

This property is essentially finding the area of a zero length interval.

Example: ∫ =8

8

0)( dxxf

This property states that if we multiply f(x) by a constant k, and then integrate, it has the same result as integrating f(x) first and then multiplying by k.

Example: If ∫ =8

1

75)(3 dxxf , then ∫ =8

1

75)(3 dxxf

This property states that if we combine two functions and then integrate the result it has the same effect as integrating each function first and then combining.

Example: If ( )∫ =±8

1

30)()( dxxgxf , then ∫ ∫ =±8

1

8

1

30)()( dxxgdxxf

This property is adding the integrals of the same function f(x) from two different intervals by essentially breaking an interval [a, c] into two subintervals [a, b] and [b, c].

Example: If ∫8

1

f

This property states that the integral of f(x) over an interval [a, b] is never smaller than the minimum value of f(x) times the length of the interval and never larger than the maximum value of f(x) times the length of the interval.

Example: If ∫12

8

f

This property states that if the function value of f(x) is greater than the function value of g(x) on an interval, the definite integral of f(x) is greater than the definite of g(x) over that interval.

Example: If (xf

Because the definite integral diagrams may be helpful to visualize some of these properties.

= 25)( dxxf and∫ =12

8

10)( dxxf , then ∫12

1

(xf

This property states that the integral of f(x) over an interval [a, b] is never smaller than the minimum value of f(x) times the length of the interval and never larger than the maximum value of f(x) times the length of the interval.

= 10)( dxxf , then max10)4()(min xf ≤≤×

This property states that if the function value of f(x) is greater than the function value of g(x) on an interval, the definite integral of f(x) is greater than the definite of g(x) over that interval.

5) =x and 2)( =xg at a point c in interval [a, b] then

dxxgdxxfb

a

b

a∫∫ ≥ )()( at the point c.

Because the definite integral is defined to be the area under a curve, the following diagrams may be helpful to visualize some of these properties.

= 35)dxx

This property states that the integral of f(x) over an interval [a, b] is never smaller than the minimum value of f(x) times the length of the interval and never larger than

)4)((xf

This property states that if the function value of f(x) is greater than the function value of g(x) on an interval, the definite integral of f(x) is greater than the definite integral

at a point c in interval [a, b] then

is defined to be the area under a curve, the following