Index FAQ

Areas of Domains and Definite Integrals

Area under a ParabolaDefinition of the Area of Certain DomainsArea under a Parabola RevisitedIntegrals and Antiderivatives

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Estimate AreasConsider the problem of determining the area of the domain bounded by the graph of the function x2, the x-axis, and the lines x=0 and x=1.

As the number n of the approximating rectangles grows, the approximation gets better.

We determine the area by approximating the domain with thin rectangles for which the area can be directly computed. Letting these rectangles get thinner, the approximation gets better and, at the limit, we get the area of the domain in question.

10

y=x2

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Estimate Areas (2)2

1

1 1The total area of the rectangles is .

n

nk

ks

n n

Height of the kth rectangle.

Length of the bottom.

Let A denote the actual area of the domain in question. Clearly sn<A for all n.

2

2 2

The approximation uses rectangles of

1 1height over the interval , .

Replace these rectangles with rectangles of height

1 to get the upper estimate

n

nk

s

k k k

n n n

k kS

n n n

1

.n

Lower est. sn

Upper est. Sn

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Estimate Areas (3)

2 2

1 1

To compute the area we have now the estimates ,

1 1 1i.e., .

n n

n n

n nk k

A s A S

k ks A S

n n n n

1

Observe that, in this case, 0.

Hence lim lim .

n n n

n nn n

S sn

S s A

2 32

1 1

To compute the limit observe that

1 1 .

n n

nk k

kS k

n n n

This can be computed directly using a previously derived formula for the sum of squares. Solution follows.

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Estimate Areas (4)3 2

2

1

Recall that .3 2 6

n

k

n n nk

32

21

Hence

1 1 1 1 1.

3 2 6 3

n

n nk

S kn n n

The blue area under the curve y=x2 over the interval [0,1] equals 1/3.

10

y=x2

Conclude

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Definition of the Area of a Domain under the Graph of a Function

Theorem

1

1

Assume that f is continuous and f 0 for . Then

1 lim min f

1 lim max f

n

nk

n

nk

x a x b

k b a k b a b ax a x a

n n n

k b a k b a b ax a x a

n n n

The previous considerations were based on some intuitive idea about areas of domain. We make that precise in the statement of the following result.

Definition

The common value of these limits is the area of the domain under the graph of the function f and over the interval [a,b].

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

The Integral (1)Theorem

1

1

Assume that f is continuous on the interval , . Then

1 lim min f

1 lim max f

n

nk

n

nk

a b

k b a k b a b ax a x a

n n n

k b a k b a b ax a x a

n n n

Definition

Notation fb

a

x dx

The common value of these limits is the integral of the function f over the interval [a,b].

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

The Integral (2)Remark

1

1

Observe that, by the previous Theorem, for a continuous function f

1lim min f

lim f f ,

where the points satisfy

n

nk

bn

knk a

k

k b a k b ax a x a

n n n

b ax x dx

n

x a

1

.k

k b a k b ax a

n n

This means that the points xk may be freely selected from the intervals [a + (k-1)/n, a + k/n]. The limit of the sum does not depend on the choice of the points xk.

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

The Integral (3)

Remark

1

Using the notation , and choosing

-- -- the points from

1the intervals

it does not matt

,

we get

er how

lim f f .

k

bn

knk a

b ax

nx

k b a k b aa a

n n

x x x dx

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Examples (1)Example 1

2

0

Compute .b

x dx

Solution

22

10

limb n

nk

bk bx dx

n n

By the Definition

3

23

1

limn

nk

bk

n3 3 2 3 3 3 3

3 2lim lim .

3 2 6 3 2 6 3n n

b n n n b b b b

n n n

Conclude

32

0

.3

b bx dx

Now use the formula for the sum of squares

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Examples (2)Example 2

2Compute .b

a

x dx

Solution

2

2 2 2

0 0

Interpreting the integral as the area

under the graph of the function

we observe that

.b b a

a

y x

x dx x dx x dx

Conclude

3 32 2 2

0 0

.3 3

b b a

a

b ax dx x dx x dx

a0 b

The red area under the graph of x2 over [a,b] equals the area over [0,b] minus the area over [0,a].

Mika Seppälä: Areas and Definite

IntegralsIndex FAQ

Integrals and Antiderivatives

3 32

3 3

x

a

x at dt

Rewriting the previous result in the following form

we observe that the integral defines the function

3 3

2F .3 3

x

a

x ax t dt

Direct differentiation yields F’(x) = x2, i.e., the function F is an antiderivative of the function f(x) = x2.

Theorem

Let f be a continuous function. The function

F f

is an antiderivative of the function f, i.e., F f .

x

a

x t dt

x x

A proof of this result will be presented later.