Author Topic: dejan91's questions (Read 17182 times) Tweet Share

dejan91 · « **Reply #45 on:** May 08, 2009, 07:36:10 pm »

If $F(x) = \int \frac{e^{2x}}{e^x + 4} dx$ and $F(0) = 3$ , find $F(x)$ .

Thanks

humph · « **Reply #46 on:** May 08, 2009, 07:47:32 pm »

Let $u = e^x + 4$ , so that $du = e^x dx$ , and so
$F(u) = \int \frac{u-4}{u}\: du = \int 1 - \frac{4}{u} \: du = u - 4 \ln u + c$ for some $c \in \mathbb{R}$ .
Thus $F(x) = e^x + 4 - 4 \ln(e^x + 4) + c$ .
Then $F(0) = 1 + 4 - 4 \ln 5 + c = 3$ , so we must have that $c = 4 \ln 5 - 2$ , and hence $F(x) = e^x + 4 \ln\left(\frac{5}{e^x + 4}\right) + 2$ .

dejan91 · « **Reply #47 on:** May 09, 2009, 12:31:50 pm »

Ahhhk ok thanks I know what I did wrong.

dejan91 · « **Reply #48 on:** May 13, 2009, 10:09:03 pm »

Am I on the right track at all?

$\int 2cos^3(3x)sin^7(3x) dx$

$= 2\int (cos(3x)sin(3x))^3sin^4(3x) dx$
$= 2\int (\frac {1}{2}sin(6x))^3sin^4(3x) dx$

I got up to this part and I'm not sure what to do from here, or even if it's correct.

Mao · « **Reply #49 on:** May 13, 2009, 10:46:48 pm »

no need to:

$\int 2\cos^3 (3x) \sin^7 (3x)\; dx = \int 2\cos(3x)(1-\sin^2 (3x)) \sin^7(dx)\; dx$

now substitute $u=\sin (3x)$

dejan91 · « **Reply #50 on:** May 13, 2009, 11:07:59 pm »

Thanks yeah I thought my way was too long.

dejan91 · « **Reply #51 on:** May 29, 2009, 04:55:31 pm »

1) Find $a$ and $b$ , such that $z = a + bi$ , and $\frac{z+i}{z+2} = i$

2) If $z = x + yi$ , find $x$ and $y$ such that $z = \sqrt{3+4i}$

Need to see if what I got was correct or not

Thanks.

Damo17 · « **Reply #52 on:** May 29, 2009, 05:35:17 pm »

Quote from: dejan91 on May 29, 2009, 04:55:31 pm

1) Find $a$ and $b$ , such that $z = a + bi$ , and $\frac{z+i}{z+2} = i$

2) If $z = x + yi$ , find $x$ and $y$ such that $z = \sqrt{3+4i}$

Need to see if what I got was correct or not Thanks.

1) $z+ i=zi+2i$
as $z=a+bi$
$a+(b+1)i=ai+bi^2+2i$
   $= -b+(a+2)i$

$\therefore$ $a=-b$ ------1
$b+1=a+2$ ------2

sub in $a=-b$ into 2
$b+1=-b+2$
   $b=\frac{1}{2}$

sub in $b=\frac{1}{2}$ into 1
$a=-\frac{1}{2}$

2) $z=\sqrt{3+4i}$ $\therefore$ $z^2=3+4i$

using the square root of z formulas:

$x^2=\frac{3+\sqrt{3^2+4^2}}{2}$
$x=\pm 2$

$y^2=x^2-a$
   $= 1$
$y=\pm 1$

therefore $z_1=2+i$ , $z_2 =-2-i$

dejan91 · « **Reply #53 on:** May 29, 2009, 05:47:40 pm »

Great

Exactly what I got...albeit using a different method.

dejan91 · « **Reply #54 on:** May 30, 2009, 12:14:05 pm »

I get the red area, but the book gets the red AND green area. The explanation at the back says that "The region 'Arg(z) is equal to or smaller than Arg(w)' is the region where a point z has 'an angle' less than $\frac{\pi}{3}$ " Why use $\frac{\pi}{3}$ ?

dcc · « **Reply #55 on:** May 30, 2009, 08:35:23 pm »

Quote from: dejan91 on May 30, 2009, 12:14:05 pm

I get the red area, but the book gets the red AND green area. The explanation at the back says that "The region 'Arg(z) is equal to or smaller than Arg(w)' is the region where a point z has 'an angle' less than $\frac{\pi}{3}$ " Why use $\frac{\pi}{3}$ ?

$Arg(w) = -\frac{2\pi}{3}$ (this is obvious from the diagram you have there). So that seems like the red area (however not as large as you have drawn it, as $|z| \leq 1$ ).

dejan91 · « **Reply #56 on:** May 31, 2009, 01:10:07 am »

Ah woops was rushing on photoshop... but yes, it was meant to be $|z| \leq 1$

yeah well that's what I thought it would be (the red area) but apparently it's not.... According to checkpoints.

kamil9876 · « **Reply #57 on:** May 31, 2009, 10:54:14 am »

Checkpoints do make mistakes, I've found a few once. They should have the exam it was taken from at the bottom so if you really want to know might as well check vcaa(if they still have it there).

dejan91 · « **Reply #58 on:** June 15, 2009, 11:23:44 pm »

To find the area underneath, do I have to find the inverse function first?
$\int_\frac{1}{2}^2log_e(2x) .dx$

TrueTears · « **Reply #59 on:** June 15, 2009, 11:26:11 pm »

Since you can not integrate $log_e(2x)$ [have to use integration by parts]

You should find the inverse and then integrate with respect to y [notice limits will change, they will be the respective y values for $\frac{1}{2}$ and 2]. Then use the whole 'rectangle' area minus the area you got from the inverse.

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