1. Not finding help here? Sign up for a free 30min 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!

Evaluating Definite Integral

  1. Jan 26, 2005 #1
    If I want to evaluate: [tex] \int^1_0 \frac {dx}{(x+1)^2} [/tex] I need to use the Fundamental Theorem of Calculus right? SO wouldn't I have to solve [tex] \int^b_a
    \frac{dx}{(x+1)^2} = \frac{(x+1)^3}{3} = \frac {8}{3} - \frac {1}{3} [/tex]? But the answer is [tex] \frac {1}{2} [/tex]
     
    Last edited: Jan 26, 2005
  2. jcsd
  3. Jan 26, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    There are a lot of errors there.How about making a substitution ??

    Daniel.

    P.S.You evaluated wrongly another integral,not the one u were supposed to...
     
  4. Jan 26, 2005 #3
    you're gona integrate that function.
    between the limits 1 and 0
    1/(x+1)^2 is the same as (x+1)^-2 {to the power neg two}
    when you integrate you add 1 to the power

    (x+1)^(-2+1) {to the power neg one}
    and then divide by this new value of the power
    (x+1)^-1
    -1
    rearanging the equation gives -1/(x+1)
    when you sub 1 for x you get -1/2
    when you sub 0 for x you get 1
    -1/2 + 1 you get 1/2
     
  5. Jan 26, 2005 #4
    I got it. It's [tex] \int^1_0 \frac {dx}{(x+1)^2} = \frac {-1}{x+1} [/tex]
    So [tex] F(1) - F(0) = \frac {1}{2} [/tex]
     
  6. Jan 26, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Not really...What about the integration limits?

    As for the limit part,could u rewrite it in an intelligible form...?

    Daniel.
     
  7. Jan 26, 2005 #6
    I am not sure what you mean. I was just using the fact that [tex] \int^b_a f(u) \ du = F(b) - F(a) [/tex]. Why wouldn't my answer be correct?

    Thanks
     
  8. Jan 26, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Because it shouldn't depend on "x"...It should be a real number...Not a function...

    Daniel.
     
  9. Jan 26, 2005 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Where did the rest of the limit go??

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Evaluating Definite Integral
  1. Evaluate integration (Replies: 6)

  2. Definite integrals (Replies: 9)

  3. Definite integrals (Replies: 6)

Loading...