MATH 136 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) gives a method of finding the “signed” area between the graph of f and the

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x-axis on the interval [

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a ,

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b]. The theorem is: FTC: If f is a continuous function on [

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a,

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b] and F is any antiderivative of f , then

Area( f , [a, b]) = F(b) − F(a) . Recall that the antiderivative of f is also called the indefinite integral denoted by f (x) dx∫ . An antiderivative is another function F such that ′ F (x) = f (x) . Because the

derivative of a constant is 0, any constant

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C may be added to F(x) to obtain another antiderivative. Thus we write f (x) dx∫ = F(x) +

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C . Then since G(x) = F(x) +

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C is also an antiderivative, the FTC states that we can also use G(x) to find Area( f , [a, b]) by computing G(b) – G(a) . However,

G(b) – G(a) = F(b) + C( ) − F(a) + C( ) = F(b) − F(a) .

Thus the additional constant

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C is irrelevant when computing Area( f , [a, b]) .

Definite Integral Notation

Because Area( f , [a, b]) is computed by using an antiderivative f (x) dx∫ = F(x) , we use the same integral sign notation to denote the signed area, but we add limits of

integration

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a and

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b to denote the interval. Thus we shall start writing f (x) dxa

b

∫ to

denote the signed area Area( f , [a, b]) . This expression is called the definite integral of f from

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a to

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b. We use the following notation to write the evaluation and computation of the definite integral:

f (x) dxa

b

∫ = F(x) ab = F(b) − F(a) .

Example 1. Evaluate the signed area between the graph of f (x) = 4 − x2 and the

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x-axis on the interval [–3, 4]. Explain the result pictorially.

Solution. We evaluate the definite integral (4 − x2 )−3

4∫ dx by finding an antiderivative

F(x) , and then computing F(4) − F(−3) :

(4 − x2 )−3

4∫ dx = 4x − 1

3x3

−3

4 = 16 − 64

3

− −12 − −27

3

=

19 − 64

3 =

−

73

units2.

Thus, the net area is –7/3 sq. units. There is more area below the

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x-axis resulting in a negative net area. −3 4

Example 2. Evaluate the signed area between the graph of f (x) = 4 sin(2x) and the

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x-axis on the interval [0, π]. Explain the result pictorially. Solution. We find an antiderivative F(x) of 4 sin(2x) and then evaluate from 0 to π by computing F(π) – F(0) :

4sin(2x ) dx0

π

∫ = 4 × 12−cos(2 x)( )

0

π= −2 cos(2x )

0π

= −2 cos(2π ) − −2 cos(0)( ) = −2 − (−2) = 0

Graphing f (x) = 4 sin(2x) on [0, π], we see that there is an equal amount of area above the

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x-axis as below the

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x-axis. Thus, the net signed area is 0. π

An definite integral can be computed numerically in two different ways on your calculator. With either way, first type the function into Y1. We shall demonstrate with

(4 − x2 )−3

4∫ dx .

The fnInt Command

On TI-84: After typing the function 4 − x2 into Y1, press Quit to return to the Home screen. Press MATH and scroll down to fnInt( and press ENTER. In general, enter the command fnInt(Y1, X, a, b). Here use the command fnInt(Y1, X, –3, 4) to obtain −2.333 . On TI-89 : After typing the function 4 − x2 into y1, press Quit to return to the Home screen. Press F3 then press 2 for ∫ ( . In general, enter the command ∫ ( y1(x), x, a, b). Here use the command ∫ ( y1(x), x, –3, 4) to obtain −2.333 . You can also use the nInt command in F3: nInt(y1(x), x,- 3, 4).

Evaluating the Integral After Graphing

On TI-84: After graphing the function 4 − x2 on the interval [–3, 4], press CALC, then press 7. Type –3 for Lower Limit and press ENTER, then type 4 for Upper Limit and press ENTER. On TI-89 or TI-92: After graphing the function 4 − x2 on the interval [–3, 4], press F5 then press 7. Type –3 for Lower Limit and press ENTER, then type 4 for Upper Limit and press ENTER.

Area Between Curves Let f ( x) and g(x ) be continuous functions over [a , b ] with f ( x) ≤ g(x ) . Then the area between the graphs of f and g is given by

Area = g( x) − f ( x)( )a

b∫ dx .

a b

f ( x)

g(x )Area

If it is not always the case that f ( x) ≤ g(x ) , then the area between the curves is

Area = g(x ) − f (x )a

b∫ dx .

Example 3. (a) Find the area between the graphs of y = 4 cos x and y = −4sin x on the interval [0, π/2]. (b) Find the area between these graphs on the interval [0, π]. Solution. (a) On [0, π/2], we have

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−4sin x ≤ 4cos x . Thus, the area between the curves is

Area = 4cos x − (−4 sin x)( )0

π /2∫ dx = 4 cos x + 4sin x( )

0

π/ 2∫ dx

= 4sin x − 4 cos x( ) 0π/ 2

= 4sin(π / 2) − 4 cos(π / 2)( ) − 4sin 0 − 4 cos 0( )

= (4 − 0) − (0 − 4) = 8 square units.

4 cos x

−4 sin x

π2

(b) On [0, π], it is not always the case that one graph is greater than the other. So we must find the point of intersection of the two graphs by solving

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4cos x = −4sin x , which gives

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−1= tan x , and x = 3π / 4 . From 0 to 3π/4, we have

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−4sin x ≤ 4cos x . But from 3π/4 to π, we have

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4cos x ≤ −4sin x . So the total area between the curves is

π2

π

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Area = 4cos x − (−4 sin x)( )0

3π / 4∫ dx + −4 sin x − 4cos x( )

3π / 4

π∫ dx

= 4cos x + 4 sin x( )0

3π / 4∫ dx + −4 sin x − 4 cos x( )

3π / 4

π∫ dx

= 4sin x − 4cos x( ) 03π / 4

+ 4cos x − 4 sin x( ) 3π / 4π

= 4sin(3π /4) − 4cos(3π /4)( ) − 4sin0 − 4cos0( )+ 4cosπ − 4sinπ( ) − 4cos(3π /4) − 4sin(3π /4)( )

=4 2

2−

4(− 2)2

− 0 − 4( )

+ 4(−1) − 0( ) − 4(− 2)2

−4( 2)

2

= 4 2 + 4 − 4 + 4 2 = 8 2 square units.

Example 4. Find the area between the graphs of k (x ) = − 1

2x + 1 and h(x ) = 1

2x2 on the

interval [–3, 3]. Solution. From the graphs of the line k (x ) and the parabola h(x ) , we see that no one function is always greater than the other over the entire interval [–3, 3].

−2 1 So we must find the points of intersection of the two graphs by solving − 1

2x + 1 = 1

2x2 .

Multiplying by 2 gives

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−x + 2 = x2 , then 0 = x2 + x − 2 = (x + 2)(x − 1). Thus, the graphs intersect at

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x = −2 and

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x =1. From –3 to –2 and from 1 to 3, we have − 1

2x + 1 ≤ 1

2x2 . But from –2 to 1, we have

12x2 ≤ −

12x + 1. Thus the area between the curves is given by

Area = k( x) − h( x)a

b∫ dx

=12x2 − −

12x + 1

−3

−2∫ dx + −

12x + 1

−

12x2

−2

1∫ dx +

12x2 − −

12x + 1

1

3∫ dx

=12x2 +

12x − 1

−3

−2∫ dx + −

12x +1 − 1

2x2

−2

1∫ dx +

12x2 +

12x −1

1

3∫ dx

=16x3 +

14x2 − x

−3

−2+ −

14x2 + x −

16x3

−2

1+

16x3 +

14x2 − x

1

3

=−86

+ 1 + 2

−

−276

+94

+ 3

+−14

+ 1 − 16

− −1 − 2 +

86

+276

+94− 3

−

16

+14− 1

=

1112

+94

+133

= 7.5 square units .