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!

Integration question

  1. Jul 1, 2016 #1
    [2][/2]1. The problem statement, all variables and given/known data
    ##∫e^x^2 + 2e^x^2x^2 dx##



    2. Relevant equations


    3. The attempt at a solution
    i let## u= x^2, ⇒ du = 2x dx, ⇒∫e^x^2 dx = e^x^2/2x ## is this correct? by using integration by parts..... i am getting ##xe^x^2-e^x^2/2x##
     
  2. jcsd
  3. Jul 1, 2016 #2

    Math_QED

    User Avatar
    Homework Helper

    You can verify whether you are correct by using wolframalpha.com. Also, your code has an error.
     
  4. Jul 1, 2016 #3
    You may use integration by parts on the first term.
    $$\int e^{x^2} \, dx=xe^{x^2}-\int x(2xe^{x^2}), dx$$
    The equation above is sufficient for your problem.
    Hope it helps.
     
  5. Jul 1, 2016 #4
    i can see you are using integration by parts, what is your ##u, du, v, dv##?
     
  6. Jul 1, 2016 #5
    I am not sure what you exactly mean by ##u,v,du,dv##. To put it more clearly,
    $$\int e^{x^2} \, dx=(e^{x^2})\int 1\ dx-\int x\frac{d(e^{x^2})}{dx} dx$$
     
  7. Jul 1, 2016 #6
    this is something "new " to me in my understanding integration by parts is ##∫udv= uv - ∫vdu## unless you have used a different approach, your method can't be described as integration by parts as stated in your opening statement.
     
  8. Jul 1, 2016 #7
    i can see you picked ## u= e^x^2, dv=1 ⇒v=x, u'=2xe^x^2 ##
     
  9. Jul 1, 2016 #8
    Thanks Abishek...........greetings from Africa
     
  10. Jul 1, 2016 #9
    i dont know when you continue with the integration of the second term the function is getting complicated i.e if you ∫2x^2e^x^2 dx or are we picking again
    dv = 1, u= 2x^2e^x^2?
     
  11. Jul 1, 2016 #10
    on ∫x.2xe^x^2 dx i let dv= x^2⇒ v=(x^3)/3, u= e^x^2 ⇒ du = 2x.e^x^2 thus
    ∫x.2xe^x^2 dx = (2e^x^2.x^3)/3-2 ∫(x^3)/3.2x.e^x^2 dx without simplification...this to me gets more complicated and no final answer is possible...
     
  12. Jul 1, 2016 #11
    lol lol lol i was blind but now i can see...

    ∫(e^x^2+2e^x^2x^2)dx= xe^x^2- ∫2e^x^2x^2dx + ∫2e^x^2x^2dx + k
    = xe^x^2 + k

    chikhabi from Africa lol
     
  13. Jul 1, 2016 #12

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    It's good to see you trying to use LaTeX, but you are having some problems.

    Is this what you mean? ##\displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\ ## It's difficult to tell with all of those run-on exponents.
     
  14. Jul 2, 2016 #13
    yes that is what i mean...how do i type that in latex?
     
  15. Jul 2, 2016 #14

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Easiest way: Hit "Reply" link under my post. See the coding.

    Otherwise, for some LaTeX here in PF that you don't such access to: Right click on the LaTeX expression. Click on "Show Math As". Click on "TeX Commands". A pop-up window should appear with the LaTeX code.

    The \displaystyle LaTeX command you see in my code gives larger format when doing "in-line" LaTeX using the ##\ \text{## ... ##}\ ## delimiters.

    Also, look up LaTeX tutorial here in Physics Forums.
     
  16. Jul 3, 2016 #15
    ##\left| \frac a b \right|## am trying practise on latex.......
    ##\left[ 1 - \left( \frac a x \right)^2 \right]^{-1/3}##
    ##\int x^2e^x \, dx##
    ##{ \displaystyle\ \int \left( e^{x^2} + 2e^{x^2}x^2 \right) dx\ }##​
     
    Last edited: Jul 4, 2016
  17. Jul 4, 2016 #16

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You can also do it in plain text, just by using parentheses correctly. When you write e^x^2 that could mean ##e^{x^2}## or ##(e^x)^2 = e^{2x}##. In plain text you would write the first as e^(x^2) and the second as (e^x)^2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integration question
  1. Integration question. (Replies: 10)

  2. Integral Questions (Replies: 2)

  3. Integration question (Replies: 2)

  4. Integral Question (Replies: 5)

  5. Integral Question (Replies: 8)

Loading...