Results 1 to 4 of 4

Math Help - Calculus Integral Problems

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    California
    Posts
    2

    Calculus Integral Problems

    Hello,

    Firstly I would like to say hello, and I'm new to these forums but I think I will use this forum very frequently

    At the moment these two problems are difficult for me to solve.

    First problem is:
    (d/dx) ∫ (0 to tanx) e^(t^2)
    Image of the problem http://i.imgur.com/D6Qh9.jpg

    Second problem is:
    the area of ∫ (-1 to 1) (x-3)/(x^2+1)
    Image of the problem http://i.imgur.com/q8EKU.jpg

    I think the answer to the first problem is e^[(tanx)^2] * (secx)^2, but unsure if that's actually correct.

    I believe the answer to the second problem Is [(1/2)(x^2)-(3x)] * arctanx? Therefore 35pi/8 + 7pi/8 = 21pi/4, but I get the feeling that this is wrong.

    Any help would be much appreciated. Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GJA
    GJA is offline
    Member
    Joined
    Jul 2012
    From
    USA
    Posts
    109
    Thanks
    29

    Re: Calculus Integral Problems

    Hi, dimelychime. Welcome to the forum!

    The first solution looks correct (this is the Fundamental Theorem of Calculus along with the chain rule).

    I got -\frac{3\pi}{2} for the second problem and checked this a couple times. Note: I used the substitution x=\tan\theta to find an antiderivative.

    I didn't see an attachment with work for the second problem so I don't know where our differences may be.

    Let me know if you try it again. If there are still differences, I will write up more details in a later post. Good luck!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    California
    Posts
    2

    Re: Calculus Integral Problems

    Hello, thanks for the reply. Yes I got -3pi/2 on my second try [last step looked (-3)(pi/4)- (-3)(-pi/4)]. I'm now unsure when to apply absolute area rule (where you would have to consider everything positive) or just integrals (hope that makes sense).

    Also, I'm pretty confident that this is the right answer to the following problem...
    ∫ {0,2} (2^x)+6 dx

    [e^ln(2)x]+6 dx

    >substitute xln(2) for u
    dx=1/ln2 du

    (1/ln2) ∫ {0,2} (e^u)+6 du
    (1/ln2) ∫ {0,2} (e^(xln2))+6 du
    (1/ln2)(2^x)+6x

    then plug in 2 and 0.

    4/ln2 - 1/ln2 = 3/ln2 <I think this is wrong
    tried it again and I got 15/ln2

    So is this right or is it unimaginably wrong?

    Thank you
    Last edited by dimelychime; August 7th 2012 at 10:35 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GJA
    GJA is offline
    Member
    Joined
    Jul 2012
    From
    USA
    Posts
    109
    Thanks
    29

    Re: Calculus Integral Problems

    To answer your first question about absolute area, I would say that it should be stated in the directions of the exercise. If I see a problem that simply says evaluate some definite integral, then I don't take absolute value. However, if the problem says something like total area then I do.

    For your second question, you're definitely on the right track. Rewriting 2^{x}=e^{x\ln 2} is a nice idea. You make the substitution u=x\ln 2 which is a legitimate technique for this integrand. There are a few things to take note of:

    1. When we make a change of variables we must also remember to change the limit of integration. For our substitution, u=x\ln 2, when x=0\Rightarrow u=0 & when x=2\Rightarrow u=2\ln 2. These now become your upper and lower limits, respectively. In other words our new definite integral to solve is

    \frac{1}{\ln 2}\int_{0}^{2\ln 2} e^{u}+6 du,

    and we don't need to go back to the x-variable.

    2. On the second line of your work you switch back to x. You don't need to do this if you follow the u-substitution all the way through. Also, even if we wanted to switch back to x we must be sure to replace du properly too. It looks as though when you switched back to x, you replaced du with dx, when really du=dx\ln 2.

    Does this help clear everything up?

    Again, nice work and good luck!
    Last edited by GJA; August 8th 2012 at 06:50 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 25th 2010, 10:41 PM
  2. Calculus Integral Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 15th 2009, 03:43 PM
  3. two calculus trig integral problems!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 20th 2008, 01:04 PM
  4. Replies: 1
    Last Post: June 23rd 2008, 09:17 AM
  5. Replies: 2
    Last Post: May 17th 2006, 05:33 AM

Search Tags


/mathhelpforum @mathhelpforum