Results 1 to 2 of 2

Thread: Help me study for my test tomorrow!

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    29

    Help me study for my test tomorrow!

    Tomorrow's my test and there are a few concepts that I have to work on.

    1. Sketch the region bounded by the graphs of the equations, and determine its area.
    $\displaystyle y=\frac{4}{\sqrt x}$
    $\displaystyle y=0, x=1, x=9$

    Answer: A=16

    My main problem with this is getting rid of the $\displaystyle \sqrt x$ from under so I can begin to integrate.

    2. Find the particular solution of the differential equation $\displaystyle f''(x)=6(x-1)$ whose graph passes through the point (2,1) and is tangent to the line $\displaystyle 3x-y-5=0$ at that point.

    I already found that $\displaystyle f'(x)=3x^2-6x$ and $\displaystyle f(x)=x^3-3x^2$. What now? I don't get what the question is asking me.

    3. Use the Second Fundamental Theorem of Calculus to find $\displaystyle F'(x)$.
    F(x)=(integration sign, x between 0 and x)$\displaystyle t^2\sqrt {1+t^3} dt$

    Answer=$\displaystyle x^2+3x+2$

    I know that $\displaystyle F(x)$ is the antiderivative of the inside of the integration symbol, but I don't understand how to find $\displaystyle F'(x)$.

    Some help on these questions would greatly be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by DarkestEvil View Post
    Tomorrow's my test and there are a few concepts that I have to work on.

    1. Sketch the region bounded by the graphs of the equations, and determine its area.
    $\displaystyle y=\frac{4}{\sqrt x}$
    $\displaystyle y=0, x=1, x=9$

    Answer: A=16

    My main problem with this is getting rid of the $\displaystyle \sqrt x$ from under so I can begin to integrate.

    rewrite as $\displaystyle \textcolor{red}{4x^{-1/2}}$ , then integrate.

    2. Find the particular solution of the differential equation $\displaystyle f''(x)=6(x-1)$ whose graph passes through the point (2,1) and is tangent to the line $\displaystyle 3x-y-5=0$ at that point.

    I already found that $\displaystyle f'(x)=3x^2-6x$ and $\displaystyle f(x)=x^3-3x^2$. What now? I don't get what the question is asking me.

    f''(x) = 6x-6

    f'(x) = 3x^2 - 6x + C ... did you calculate the constant of integration?
    what information did they give you that might help you find it?

    3. Use the Second Fundamental Theorem of Calculus to find $\displaystyle F'(x)$.
    F(x)=(integration sign, x between 0 and x)$\displaystyle t^2\sqrt {1+t^3} dt$

    Answer=$\displaystyle x^2+3x+2$

    I know that $\displaystyle F(x)$ is the antiderivative of the inside of the integration symbol, but I don't understand how to find $\displaystyle F'(x)$.

    your "answer" is incorrect.

    $\displaystyle \textcolor{red}{\frac{d}{dx} \int_a^x f(t) \, dt = f(x)}$
    ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Aug 31st 2010, 11:38 AM
  2. Replies: 2
    Last Post: Oct 3rd 2009, 02:30 PM
  3. Replies: 1
    Last Post: Sep 14th 2009, 11:18 PM
  4. Help with test study
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 10th 2008, 08:03 PM
  5. Test Study Help
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: Oct 25th 2007, 08:04 PM

Search Tags


/mathhelpforum @mathhelpforum