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

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

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

$\displaystyle -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx$$\displaystyle = -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx + A by part B. \displaystyle = -\int^{b}_{a} f(x) dx + A + A Middle integral is equal to \displaystyle A is given in the question Finally: \displaystyle -\int^{b}_{a} f(x) dx=\int^{a}_{b} f(x) dx=$$\displaystyle \int^{a}_{0} f(x) dx-\int^{b}_{0} f(x) dx=A-0=A$.

So:

$\displaystyle -\int^{b}_{a} f(x) dx + \int^{a}_{0} f(x) dx - \int^{-a}_{0} f(x) dx=3A$.

RonL

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

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

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

$\displaystyle \int^{b}_{a} f(x) dx=\int^{b}_{0} f(x) dx-\int^{a}_{0} f(x) dx =0-A=-A$

as the two integrals on the RHS are given,

and:

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

is given.

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

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

RonL

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

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

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

RonL

4. 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.
$\displaystyle \frac{dQ}{dt} = \frac{-5Q}{5000-4t}$

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

5. Originally Posted by CaptainBlack
Looks like making the substitution $\displaystyle \cos(u)=x$, and then
integrating by parts.

RonL
I think the question is asking for:

$\displaystyle \int_0^a \arccos(x)\ dx$.

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

$\displaystyle -\int_{\pi/2}^{\arccos(a)} u\ \sin(u)\ du$.

Then proceed with integration by parts.

RonL

6. 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.
$\displaystyle \frac{dQ}{dt} = \frac{-5Q}{5000-4t}$

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

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

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

RonL

7. thanks very much

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