1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    ∫(x/3x- x/3x^.5)dx

    =[((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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook