Integrals - Math148

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    To find the area under the graph of a nonnegative,continuous function f over the interval [ a , b]:

    1. Find any antiderivative F ( x) of f ( x). (The simplestis the one for which the constant of integration is 0.)

    2. Evaluate F ( x) using b and a , and compute

    F (b) F (a ). The result is the area under the graphover the interval [ a , b].

    4.3 Area and Definite Integrals

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    Example 2: Find the area under the graph of y = x2 +1 over the interval [ 1, 2].

    1. Find any antiderivative F ( x) of f ( x). We choose thesimplest one.

    4.3 Area and Definite Integrals

    F ( x) x3

    3

    x

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    Example 2 (concluded): 2. Substitute 2 and 1, and find the difference

    F (2) F ( 1).

    4.3 Area and Definite Integrals

    F (2) F ( 1) 23

    32 ( 1)

    3

    3( 1)

    8

    3 2 1

    31

    83

    2 13

    1

    6

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    DEFINITION:

    Let f be any continuous function over the interval[a , b] and F be any antiderivative of f . Then, thedefinite integral of f from a to b is

    4.3 Area and Definite Integrals

    f ( x) dxa

    b F (b) F (a ).

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    Example 4: Evaluate

    Using the antiderivative F ( x) = x3/3, we have

    It is convenient to use an intermediate notation:

    where F ( x) is an antiderivative of f ( x).

    4.3 Area and Definite Integrals

    x2dxab

    .

    x2dxa

    b b3

    3 a

    3

    3.

    f ( x) dxa

    b F ( x) ab F (b) F (a ),

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    Example 5: Evaluate each of the following:

    4.3 Area and Definite Integrals

    a) ( x2 x)1

    4 dx;

    b) e x0

    3 dx;

    c) 1 2 x 1

    x

    1

    e dx (assume x 0).

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    Example 5 (continued):

    4.3 Area and Definite Integrals

    a) ( x2 x)1

    4 dx x

    3

    3 x2

    21

    4

    4 3

    3 4 2

    2

    ( 1)3

    3 ( 1) 2

    2

    643

    162 13 12

    643

    8 13

    12

    14 16

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    Example 5 (continued):

    4.3 Area and Definite Integrals

    b) e x dx0

    3 e x

    0

    3 e3 e0

    e3 1

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    Example 5 (concluded):

    4.3 Area and Definite Integrals

    c) 1 2 x 1 x

    1

    e dx x x2 ln x

    1

    e

    (e e2 ln e) (1 12 ln1)

    (e e2 1) (1 1 0)

    e e2 1 1 1

    e e2 3

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    THE FUNDAMENTAL THEOREM OFINTEGRAL CALCULUS

    If a continuous function f has an antiderivative F over[a , b], then

    4.3 Area and Definite Integrals

    limn

    f ( xi) x

    i 1

    n

    f ( x) d a

    b x F (b) F (a ).

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    Example 6: Suppose that y is the profit per mile traveled and x is number of miles traveled, in thousands. Find the area under y = 1/ x over the interval [1, 4] and interpret the significance ofthis area.

    4.3 Area and Definite Integrals

    dx x1

    4 ln x 14

    ln 4 ln1 ln 4 0 1.3863

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    Example 6 (concluded):The area represents a total profit of $1386.30 when themiles traveled increase from 1000 to 4000 miles.

    4.3 Area and Definite Integrals

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    Example 8: Predict thesign of the integral by, using area, and then evaluatethe integral.

    From the graph, it appearsthat there is considerablymore area below the x-axisthan above. Thus, we expectthat the sign of the integralwill be negative.

    4.3 Area and Definite Integrals

    Consider ( x3 3 x 1)dx1

    2 .

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    Example 9: Northeast Airlines determines that themarginal profit resulting from the sale of x seats on a

    jet traveling from Atlanta to Kansas City, in hundredsof dollars, is given by

    Find the total profit when 60 seats are sold.

    4.3 Area and Definite Integrals

    P ( x) x 6.

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    Example 9 (continued):We integrate to find P (60).

    4.3 Area and Definite Integrals

    P 60 P x dx0

    60

    x 6 dx060 2

    3

    x3 2 6 x0

    60

    23

    60 3 2 6 60

    23

    0 3 2 6 0

    50.1613

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    Example 9 (concluded):When 60 seats are sold, Northeasts profit is

    $5016.13. That is, the airline will lose $5016.13 on

    the flight.

    4.3 Area and Definite Integrals

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    Example 12: A particle starts out from some origin.Its velocity, in miles per hour, is given by

    where t is the number of hours since the particle leftthe origin. How far does the particle travel during thesecond, third, and fourth hours (from t = 1 to t = 4)?

    4.3 Area and Definite Integrals

    v(t ) t t ,

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    Example 12 (continued):Recall that velocity, or speed, is the rate of change ofdistance with respect to time. In other words, velocity

    is the derivative of the distance function, and thedistance function is an antiderivative of the velocityfunction. To find the total distance traveled from t = 1

    to t = 4, we evaluate the integral

    4.3 Area and Definite Integrals

    t t dt 1

    4 .

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    Example 12 (concluded):

    4.3 Area and Definite Integrals

    t t dt 1

    4 t 1 2 t dt

    1

    4

    23 t

    3 2 12 t

    2

    1

    4

    23

    4 3 2 12

    4 2 2

    313 2

    12

    12

    163

    162

    23

    12

    143

    152

    736

    121

    6 mi.