How to find the area under a curve (quotient integral?)?

For HW I need to find the area of the region under the curve y=(4x + 4)/(x-1) from a=2 to b=3.

I don't really know where to begin with this problem, and my textbook isn't much help. Thanks in advance for any help you can offer!

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  • Ian H
    Lv 7
    6 years ago
    Favorite Answer

    4x + 4 = 4x - 4 + 8,

    so that shows you that you can split the integrand as

    (4x + 4)/(x - 1) = 4 + 8/(x - 1) (see note below)

    I = {x: 2 to 3} ∫[(4x + 4)/(x - 1)]dx = |4x + 8ln(x - 1)| from 2 to 3

    I = 12 + 8ln(2) - 8 - 8ln(1) = 8ln(2) + 4 ~ 9.545 sq units

    Note: Splitting quotients of various kinds into fractions that are each easier to integrate is a topic called "partial fractions" and there are a few methods and rules to know.

    This website gives an understandable and full explanation of PF's with examples

    http://math.ucsd.edu/~wgarner/math10b/partialfract...

    The denominator factors relate to the form of partial fractions of 4 main types :-

    a) Partial Fractions with Distinct Linear Factors

    b) Partial Fractions with Some Repeated Linear Factors

    c) Partial Fractions with Distinct Irreducible Quadratic Factors

    d) Partial Fractions with Some Repeated Irreducible Quadratic Factors

    Regards - Ian H

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  • 6 years ago

    int [ ( 4x + 4 ) / ( x - 1 ) dx ]

    = int [ 4 + 8 / ( x - 1 ) dx ]

    = 4x + 8 ln abs ( x - 1 ) + C

    Source(s): my brain
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