Section 5.2The Definite Integral

(1) The Definite Integral and Net Area(2) Properties of the Definite Integral(3) Evaluating Definite Integrals using Limits

The Definite Integral of a FunctionThe definite integral of a function f on the interval [a,b] is∫ b

af (x)dx = lim

n→∞n∑

i=1f (xi )∆x

(where ∆x = b−an and xi = a+ i∆x), provided that this limit exists.

The integral symbol∫

is an elongated S , introduced by Leibniz because

an integral is a limit of sums.

The procedure of calculating an integral is integration.f (x) is the integrand and {a,b} are the limits of integration.

If∫ b

af (x)dx is defined, we say that f is integrable on [a,b].

If f is continuous on [a,b], or if f has only a finite number ofjump discontinuities, then f is integrable on [a,b].

x

y

y = f (x)

a b

+ +–∑

f (xi )∆x is an approximation of the net area.

x

y

y = f (x)

a b

+ +–∫ b

af (x)dx is the net area.

The definite integral calculates net area: the area below the positivepart of a graph minus the area above the negative part.

x

y

y = f (x)

-2 -1 1 2

1

2

3Example 1: Evaluate

∫ 2

−2f (x)dx using geometry.

Solution: The left half is a quarter-circle of radius 2and the right is a 2×2 rectangle below a base 2,height 1 triangle. The area is 5+π.

Net Area and Total Area

Example 2: For the function f (x) shown to

the right, evaluate∫ 3

−2f (x)dx .

BG

R

x

y

y= f(x)

-2 -1 1 2 3

-2

-1

1

2

Solution:∫ 3

−2f (x)dx =B +G −R = π+ 1

2−2 = π− 3

2.

In contrast, the total area contained between the graph and the x-axis is∫ 3

−2

∣∣f (x)∣∣ dx =B +G +R =π+ 52

.

Properties of the Definite Integral

Integral Property #1∫ b

af (x)dx =−

∫ a

bf (x)dx

For the first integral, the base of each rectangle in the Riemann sum haslength b−a

n , while in the second integral each rectangle has basea−bn =− b−a

n .

Integral Property #2∫ a

af (x)dx = 0

The interval [a,a] has length 0, so the region above or below the graph isjust a line segment, which has area 0.

Properties of the Definite Integral

Integral Property #3∫ b

ac dx = c(b−a)

If f (x)= c , then the area below the graph of f is a rectangle with baseb−a, height c , and therefore area c(b−a).

Integral Property #4∫ b

af (x)dx =

∫ c

af (x)dx +

∫ b

cf (x)dx

Properties of the Definite Integral

Integral Property #5∫ b

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

∫ b

af (x)dx +

∫ b

ag(x)dx

x

y

A

A= f (xi )∆x

a bxi xi+1

y = f (x)

x

y

B

B = g(xi )∆x

a bxi xi+1

y = g(x)

x

y

C

C = (f (xi )+g(xi ))∆x

a bxi xi+1

y = f (x)+g(x)

The brown area is the sum of the blue and the red area.

∫ b

a(f (x)−g(x)) dx =

∫ b

af (x)dx −

∫ b

ag(x)dx

Properties of the Definite Integral

Integral Property #6∫ b

a(cf (x)) dx = c

∫ b

af (x)dx for any constant c .

Stretching a graph vertically by a factor of c multiplies net area by c .

x

y

A

A= f (xi )∆x

a bxi xi+1

y = f (x)

x

y

B

B =2f (xi )∆x

a bxi xi+1

y =2f (x)

The red area is twice as the blue area.

Integral Comparison Property #1

If f (x)≥ 0 for a≤ x ≤ b, then∫ b

af (x)dx ≥ 0.

(In this case, net area just means total area under the curve.)

Integral Comparison Property #2

If f (x)≥ g(x) on the interval [a,b], then∫ b

af (x)dx ≥

∫ b

ag(x)dx .

x

y y = f (x)

y = g(x)

a b

Integral Comparison Property #3If m≤ f (x)≤M on the interval [a,b], then

m(b−a)≤∫ b

af (x)dx ≤M(b−a).

x

y

y =m

y = f (x)

y =M

a b

Evaluating Definite Integrals as LimitsThe definite integral of f on the interval [a,b] is∫ b

af (x)dx = lim

n→∞n∑

i=1f (xi )∆x

where ∆x = b−a

nand xi = a+ i∆x , provided that this limit exists.

Example 3: Express the definite integral∫ 5

1

2x1−x3 dx

as a limit of Riemann sums.

Answer: limn→∞

n∑i=1

2(1+ 4i

n

)1−

(1+ 4i

n

)3 4n

Example 4: What definite integral is represented by

limn→∞

n∑i=1

((2+ 3i

n

)sin

(2+ 3i

n

))3n? Answer:

∫ 5

2x sin(x)dx

Example 5: Evaluate the integral∫ 3

1

(x2−6

)dx .

∆x = b−a

n= 2n

xi = a+ i∆x = 1+ 2in

∫ 3

1

(x2−6

)dx = lim

n→∞n∑

i=1f (xi )∆x

= limn→∞

n∑i=1

((1+ 2i

n

)2−6

)2n

= limn→∞

n∑i=1

(8in2 + 8i2

n3 − 10n

)

= limn→∞

(8n2

n(n+1)2

+ 8n3

n(n+1)(2n+1)6

− 10nn

)= 10

3