# Problems with some integration

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• Jul 22nd 2006, 09:04 PM
CaptainBlack
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:

$
-\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
$

by part B.

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

Finally:

$
-\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
$
.

So:

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

RonL
• Jul 22nd 2006, 09:21 PM
CaptainBlack
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:

$
\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:

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

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
• Jul 22nd 2006, 09:23 PM
CaptainBlack
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
• Jul 23rd 2006, 10:07 PM
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 | <br /> img.top {vertical-align:15%;}<br /> $t_n$ | <br /> img.top {vertical-align:15%;}<br /> $Q_n$  | ------------------ | 0 | 0  | 12500 | | 1 |    |      | ... | 5 |    |      |
Any suggestions?
• Jul 24th 2006, 04:29 AM
CaptainBlack
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:

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

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

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

Then proceed with integration by parts.

RonL
• Jul 24th 2006, 04:47 AM
CaptainBlack
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 | <br /> img.top {vertical-align:15%;}<br /> $t_n$ | <br /> img.top {vertical-align:15%;}<br /> $Q_n$  | ------------------ | 0 | 0  | 12500 | | 1 |    |      | ... | 5 |    |      |
Any suggestions?

You need a table like:

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

RonL
• Jul 25th 2006, 03:16 AM
scorpion007
thanks very much
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