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Math Help - Problems with some integration

  1. #16
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    Quote Originally Posted by scorpion007
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    C: -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx = 3A(
    Is true:

    <br />
-\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx  = -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx + A<br />

    by part B.

    <br />
= -\int^{b}_{a} f(x) dx + A + A<br />
    Middle integral is equal to A is given in the question

    Finally:

    <br />
 -\int^{b}_{a} f(x) dx=\int^{a}_{b} f(x) dx= \int^{a}_{0} f(x) dx-\int^{b}_{0} f(x) dx=A-0=A<br />
.

    So:

    <br />
-\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx=3A<br />
.

    RonL
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  2. #17
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    Quote Originally Posted by scorpion007
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    D: \Bigl\vert \int^{b}_{a} f(x) dx \Bigr\vert + \Bigl\vert \int^{a}_{0} f(x) dx \Bigr\vert = 2A
    Is true, as:

    <br />
\int^{b}_{a} f(x) dx=\int^{b}_{0} f(x) dx-\int^{a}_{0} f(x) dx =0-A=-A<br />

    as the two integrals on the RHS are given,

    and:

    \int^{a}_{0} f(x) dx=A

    is given.

    But as f(x) \ge 0 on [0,a], A \ge 0 so:

    \Bigl\vert \int^{b}_{a} f(x) dx \Bigr\vert + \Bigl\vert \int^{a}_{0} f(x) dx \Bigr\vert =|-A|+|A|= 2A

    RonL
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  3. #18
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    Quote Originally Posted by scorpion007
    Here's also another question i couldnt solve. Its multiple choice. See if you can get it.

    f is a continuous, differentiable function with f(x) = f(-x) such that f(x) \ge 0 for x \in [0, a] and f(x) < 0 for x \in (a, b]. It is also known that \int^{b}_{0} f(x) dx = 0 and \int^{a}_{0} f(x) dx = A. Which of the following is false?

    E: \int^{-a}_{-b} f(x) dx = 0
    Is obviously false.

    RonL
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  4. #19
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    thanks a lot!

    I have another question:

    Given that there is 12.5 kg (12500 g) or chlorine at time t=0 in a pool, use the Euler's method of approximation with h=30 to solve this differential equation:

    Q is amount of chlorine in grams
    t is in minutes.
    \frac{dQ}{dt} = \frac{-5Q}{5000-4t}

    Basically I have to fill a table like so:
    Code:
    | n | 
    
    
    
    
    t_n | 
    
    
    
    
    Q_n  |
    ------------------
    | 0 | 0  | 12500 |
    | 1 |    |       |
    ...
    | 5 |    |       |
    Any suggestions?
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  5. #20
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    Quote Originally Posted by CaptainBlack
    Looks like making the substitution \cos(u)=x, and then
    integrating by parts.

    RonL
    I think the question is asking for:

    <br />
\int_0^a \arccos(x)\ dx<br />
.

    Let \cos(u)=x, then -\sin(u)\ \frac{du}{dx}=1 so the integral becomes:

    <br />
-\int_{\pi/2}^{\arccos(a)} u\ \sin(u)\ du <br />
.

    Then proceed with integration by parts.

    RonL
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  6. #21
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    Quote Originally Posted by scorpion007
    thanks a lot!

    I have another question:

    Given that there is 12.5 kg (12500 g) or chlorine at time t=0 in a pool, use the Euler's method of approximation with h=30 to solve this differential equation:

    Q is amount of chlorine in grams
    t is in minutes.
    \frac{dQ}{dt} = \frac{-5Q}{5000-4t}

    Basically I have to fill a table like so:
    Code:
    | n | 
    
    
    
    
    t_n | 
    
    
    
    
    Q_n  |
    ------------------
    | 0 | 0  | 12500 |
    | 1 |    |       |
    ...
    | 5 |    |       |
    Any suggestions?
    You need a table like:

    <br />
\begin{array}{c|ccc}<br />
n&t_n&Q'_n&Q_n\\ \hline<br />
0&0&-12.5&12500\\<br />
1&30&-12.81&12125\\<br />
:&:&:&:<br />
\end{array}<br />

    where Q_n=Q_{n-1}+h\ Q'_{n-1}, Q'_n=\frac{-5Q_n}{5000-4t_n}, and t_n=n.h=n.30

    RonL
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  7. #22
    Senior Member
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    thanks very much
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