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Homework Help: Basic Definite Integration, area under the curve

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate:

    A:
    [tex]
    \int^{16}_0 (\sqrt{x} - 1)\,dx
    [/tex]

    B:
    [tex]
    \int^4_1 \frac{2}{\sqrt{x}}\,dx
    [/tex]

    C:
    [tex]
    \int^6_2 \frac{4}{x^3}\,dx
    [/tex]


    2. Relevant equations

    n/a

    3. The attempt at a solution
    Its part C which I think I have done wrong, but the others could be too.

    Part A:
    [tex]
    \int^{16}_0 (\sqrt{x} - 1)\,dx = \int^{16}_0 (x^{\frac{1}{2}}-1)\,dx = [\frac{2}{3}x^{1.5}-1x]^{16}_0 = [112]-[0] = 112 \,sq\, units
    [/tex]

    Part B:
    [tex]
    \int^4_1 \frac{2}{\sqrt{x}}\,dx = \int^4_1 2x^{-\frac{1}{2}}\,dx = [4x^{\frac{1}{2}}]^4_1 = [8] - [4] = 4 \,sq\, units
    [/tex]

    C:
    [tex]
    \int^6_2 \frac{4}{x^3}\,dx = \int^6_2 4x^{-3}\,dx = [-2x^{-2}]^6_2 = [-\frac{1}{18}] - [-\frac{1}{2}] = 0.44 \,sq\, units
    [/tex]
     
    Last edited: May 2, 2013
  2. jcsd
  3. May 2, 2013 #2
    first is incorrect - others are fine (assuming you mean 4/9 for the third).

    apply the limits again
     
  4. May 2, 2013 #3
    Ah yeah, dont know what i was thinking there!

    Should it be...
    [tex]
    [\frac{80}{3}]-[0] = \frac{80}{3} sq\,units
    [/tex]

    And yeah 4/9 is what I meant its just the way it came out on my calc I couldnt work out what fraction produced that recurring decimal :)

    Thanks for your help.
     
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