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Using integration to find the area between two lines

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the area under (x/3x) and above (x/3x^.5) between x=1 and x=4.


    2. Relevant equations
    -Area of a representative rectangle= ((x/3x)-(x/3x^.5))dx
    -To integrate, raise the power of part of an expression and then divide the number in front by the raised exponent.

    3. The attempt at a solution
    I'm not the best at simplifying but here it goes:
    4
    ∫(x/3x- x/3x^.5)dx
    1


    =[((x/2)^2) / (3/2x^2) - ((x/2^2) / (3/1.5)x^1.5] between 1 and 4


    From there the answers are not reasonable (too low). Any help?
     
    Last edited: Jun 20, 2012
  2. jcsd
  3. Jun 20, 2012 #2

    LCKurtz

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    Is that ##\frac x {3x}## or ##(\frac x 3) x##? Ditto for the second one. And in either case you should simplify before integrating.
    And that wouldn't be how you integrate fractions. Do the algebra first.
     
  4. Jun 21, 2012 #3
    Like LCKurtz said, the way you write it is ambiguous. Write it in ##\LaTeX## and then just subtract one integral from the other. Either case, your integrals are wrong. What you did was make the terrible error that ##\int\frac{f(x)}{g(x)}\mbox{d}x=\frac{\int f(x)\mbox{d}x}{\int g(x)\mbox{d}x}## which is obviously wrong.
     
  5. Jun 21, 2012 #4

    HallsofIvy

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    As the others have said, this does not really make sense. It would be more reasonable to write the first as simply 1/3 and the second as x^.5/3. If you cannot us LaTeX at least use parentheses to make clear what you mean.

    Yes, the integral of [itex]x^n[/itex] is [itex]x^{n+1}/(n+1)[/itex] but you cannot just apply that to parts of an integral, in particular to the numerator and denominator of a fraction. Remember the "product rule" and "quotient rule" for differentiation? You did not just differentiate the parts of a product or quotient and cannot do that for integration.

    [tex]\int_1^4 ((1/3)- (1/3)x^{.5})dx= (1/3)\int_1^4 (1- x^{.5})dx][/tex]
    [tex]= \left[x- (1/1.5)x^{1.5}\right]_1^4[/tex]


     
  6. Jun 21, 2012 #5
    Thank you everyone so much! I forgot all about the product and quotient rule. I'm sorry for all of my mistakes, I just started learning calculus. Thanks again.
     
  7. Jun 21, 2012 #6
    You're absolutely right but I think I was saying that I would integrate everything as x=4 and then subtract integrating it as x=1 to get my answer.
     
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