Results 1 to 5 of 5

Math Help - Integration

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    76

    Integration

    Hi guys, im given this question:

    With the given equation 2^t + t^2, find the area between 0 and 1 via:

    i) finding certain limits.

    ii) Fundamental theorem of calculus.

    Can someone show me what to do for part i). I really dont know what to do.

    For part ii, i was wondering if this is correct: Integrate 2^t and t^2. That becomes 2^t = 2^t / ln2 + t^3/3 and then put in the values of 0 and 1 respectively. Is it the right way?

    Thanks guys
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    If you consider a definite integral as limit of a 'Riemann sum' is...

    \int_{0}^{1} f(t)\cdot dt = \lim_{n \rightarrow \infty} \frac{1}{n}\cdot \sum_{i=1}^{n} f(\frac{i}{n}) (1)

    In our case is f(t)= 2^{t} + t^{2} so that...

    f(\frac{i}{n}) = 2^{\frac{i}{n}} + (\frac{i}{n})^{2} (2)

    Now is...

    \sum_{i=1}^{n} 2^{\frac{i}{n}} = \frac{2-2^{\frac{1}{n}}}{2^{\frac{1}{n}}-1}

    \sum_{i=1}^{n}(\frac{i}{n})^{2}= \frac{1}{n^{2}}\cdot \sum_{i=1}^{n} i^{2} = \frac {n\cdot (n+1)\cdot (2n+1)}{6\cdot n^{2}} (3)

    ... and...

    2^{\frac{1}{n}}-1= e^{\frac{\ln 2}{n}}-1 = \frac{\ln 2}{n} + \frac{\ln^{2} 2}{2\cdot n^{2}} + \dots (4)

    ... so that (1) becomes...

    \int_{0}^{1} (2^{t} + t^{2})\cdot dt = \lim_{n \rightarrow \infty} \frac{1}{n}\cdot (\frac {2-2^{\frac{1}{n}}}{2^{\frac{1}{n}}-1} + \frac{2\cdot n^{3} + 3\cdot n^{2} + n}{6\cdot n^{2}}) = \frac {1}{\ln 2} + \frac{1}{3} (5)

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 6th 2009 at 07:10 AM. Reason: a previous error... now corrected...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2009
    Posts
    76
    Thanks a lot chisigma! So the Riemann sum is used to find certain limits?

    As well, i was wondering in equation 3 how did u derive the numerator and denominator?

    Thank you sooo much!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    In the passage (3) you have to use the known result...

    \sum_{i=1}^{n} i^{2} = \frac{n\cdot (n+1)\cdot (2n+1)}{6}

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2009
    Posts
    76
    Quote Originally Posted by chisigma View Post
    In the passage (3) you have to use the known result...

    \sum_{i=1}^{n} i^{2} = \frac{n\cdot (n+1)\cdot (2n+1)}{6}

    Kind regards

    \chi \sigma
    thanks again chisigma


    Much appreciated
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 01:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 05:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 11:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 07:58 AM

Search Tags


/mathhelpforum @mathhelpforum