5.3 Definite Integrals and Antiderivatives

Use graphs and your knowledge of area and

x3

0

1

∫ dx = 14

to evaluate the integral.

a) x3

−1

1

∫ dx b) x3 + 3( )0

1

∫ dx

c) x - 2( )3

2

3

∫ dx d) x3

−1

1

∫ dx

d) 1 - x3( )0

1

∫ dx e) x - 1( )3

−1

2

∫ dx

0 13

4

1

4

1

2

3

4−

1

4

Use graphs and your knowledge of area and

x3

0

1

∫ dx = 14

to evaluate the integral.

g) x2

⎛⎝⎜

⎞⎠⎟3

0

2

∫ dx h) x3

−8

8

∫ dx

i) x3 - 1( )0

1

∫ dx j) x3

0

1

∫ dx

01

2

−3

43

4

Page 285 gives rules for working with integrals, the most important of which are:

2. ( ) 0a

af x dx =∫ If the upper and lower limits are equal,

then the integral is zero.

1. ( ) ( )b a

a bf x dx f x dx= −∫ ∫ Reversing the limits

changes the sign.

( ) ( )b b

a ak f x dx k f x dx⋅ =∫ ∫3. Constant multiples can be

moved outside.

→

1.

( ) 0a

af x dx =∫ If the upper and lower limits are equal,

then the integral is zero.2.

( ) ( )b a

a bf x dx f x dx= −∫ ∫ Reversing the limits

changes the sign.

( ) ( )b b

a ak f x dx k f x dx⋅ =∫ ∫3. Constant multiples can be

moved outside.

( ) ( ) ( ) ( )b b b

a a af x g x dx f x dx g x dx+ = +⎡ ⎤⎣ ⎦∫ ∫ ∫4.

Integrals can be added and subtracted.

→

( ) ( ) ( ) ( )b b b

a a af x g x dx f x dx g x dx+ = +⎡ ⎤⎣ ⎦∫ ∫ ∫4.

Integrals can be added and subtracted.

5. ( ) ( ) ( )b c c

a b af x dx f x dx f x dx+ =∫ ∫ ∫

Intervals can be added(or subtracted.)

a b c

( )y f x=

→

Ex: Suppose f and h are continuous functions such that:

f x( )1

9

∫ dx = -1, f x( )7

9

∫ dx =5, h x( )7

9

∫ dx = 4

a) −2 f x( )1

9

∫ dx

b) f x( ) + h(x) ( ) 7

9

∫ dx

c) 2 f x( )−3 h(x)( )7

9

∫ dx

2

9

-2

Ex: Suppose f and h are continuous functions such that:

f x( )1

9

∫ dx = -1, f x( )7

9

∫ dx =5, h x( )7

9

∫ dx = 4

d) f (x)9

1

∫ dx

e) f(x)1

7

∫ dx

f) h x( )− f(x)( )9

7

∫ dx

1

-6

1

The average value of a function is the value that would give the same area if the function was a constant:

0

1

2

3

4

5

1 2 3

21

2y x=

3 2

0

1

2A x dx=∫

33

0

1

6x=

27

6=

9

2= 4.5=

4.5Average Value 1.5

3= =

1.5

→

Exploration Activity

How long is the average chord of a circle with radius r?

1

-1

-2 2-r r

favg =1

b−ar2 −x2

a

b

∫ dx

favg =1

r−(−r)2 r2 −x2

−r

r

∫ dx

favg =12r

2 r2 −x2

−r

r

∫ dx

favg =12r

2πr2

2 =

πr2

The mean value theorem for definite integrals says that for a continuous function, at some point on the interval the actual value will equal to the average value.

Mean Value Theorem (for definite integrals)

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

( ) ( )1

b

af c f x dx

b a=

− ∫

π