Logs (differentiation + integration) (1 Viewer)

Makro · Jun 7, 2009

I can't figure out how to do this. I did this a month ago and now I'm revising and just can't get it. It's really killing me.

5. e) Differentiate: ln (7x - 2)
5. g) differentiate: log₂ x

6. d) Find the integral of: (e^2x + 5) limits being 1 to 0.
6. e) '' xe^x² (this one is indefinite)

Would someone explain the process of integrating with logs, my textbook/workbook doesn't go into enough detail. I really need baby steps here!

kurt.physics · Jun 7, 2009

Makro said:
I can't figure out how to do this. I did this a month ago and now I'm revising and just can't get it. It's really killing me.

5. e) Differentiate: ln (7x - 2)
5. g) differentiate: log₂ x

6. d) Find the integral of: (e^2x + 5) limits being 1 to 0.
6. e) '' xe^x² (this one is indefinite)

Would someone explain the process of integrating with logs, my textbook/workbook doesn't go into enough detail. I really need baby steps here!

$$Remembering that ln is the same as$\,\,log_{e}$

$5 e) \frac{d}{dx}\left(log_{e}\left(7x -2\right) \right)$

$$If$\,\,log_{e}f(x)\,\,$then$\,\, \frac{dy}{dx} = f'(x) \cdot \frac{1}{f(x)}$

$f(x) = 7x - 2$

$f'(x) = 7$

$\therefore \frac{d}{dx}\left(log_{e}\left(7x -2\right) \right) = 7 \cdot \frac{1}{7x-2}$

$= \frac{7}{7x-2}$

$$5 g)$\,\,\frac{d}{dx}\left( log_{2}\,\,x\,\,\right)$

$Using change of base$

$log_{2}\,\,x\,\, = \frac{ log_{e}\,\,x\,\,}{ log_{e}\,\,2\,\,}$

$note that$\,\,log_{e}\,\,2\,\,$is just a number like 5 or -1, ie a constant$

$\therefore \frac{d}{dx}\left( log_{2}\,\,x\,\,\right) = \frac{d}{dx}\left( \frac{ log_{e}\,\,x\,\,}{ log_{e}\,\,2\,\,}\right)$

$=\frac{1}{log_{e}\,\,2\,\,} \cdot \frac{d}{dx}\left(log_{e}\,\,x\,\,\right)$

$\frac{d}{dx}\left(log_{e}\,\,x\,\,\right) = \frac{1}{x}$

$= \frac{1}{log_{e}\,\,2\,\,}\cdot \frac{1}{x}$

$= \frac{1}{x\,log_{e}\,\,2\,\,}$

$\int_{0}^{1} \left(e^{2x} + 5\right) dx$

$\int e^{ax+b} dx = \frac{1}{a}e^{ax+b}$

$\therefore \int_{0}^{1} \left(e^{2x} + 5\right) dx = \left[ \frac{1}{2}e^{2x} + 5x\right]_{0}^{1}$

$= \left( \frac{1}{2}e^{2 \cdot 1} + 5 \cdot 1\right) - \left( \frac{1}{2}e^{2 \cdot 0} + 5 \cdot 0 \right)$

$= \frac{1}{2}e^{2} + 5 - \frac{1}{2}e^{0}$

$= \frac{1}{2}e^2 + 5 - \frac{1}{2}$

$= \frac{1}{2}e^2 + 4 \frac{1}{2}$

$\int xe^{x^2} dx$

$= \int \frac{1}{2} \cdot 2xe^{x^2} dx$

This is in the form:

$f'(x) e^{f(x)}$

Now,

$$If$\,\,y = e^{f(x)}\,\,$then$\,\,\frac{dy}{dx} = f'(x) e^{f(x)}$

So in reverse,

$\int f'(x) e^{f(x)} = e^{f(x)}$

So we may now proceed as such:

$\int \frac{1}{2} \cdot 2xe^{x^2} dx = \frac{1}{2} \int 2xe^{x^2} dx$

$= \frac{1}{2} \cdot e^{x^2} + C$

$= \frac{1}{2} e^{x^2} + C$

zazzy1234 · Jun 7, 2009

Makro said:
I can't figure out how to do this. I did this a month ago and now I'm revising and just can't get it. It's really killing me.

5. e) Differentiate: ln (7x - 2)
5. g) differentiate: log₂ x

6. d) Find the integral of: (e^2x + 5) limits being 1 to 0.
6. e) '' xe^x² (this one is indefinite)

Would someone explain the process of integrating with logs, my textbook/workbook doesn't go into enough detail. I really need baby steps here!

a. with such question u jsut differeniate

"7x-2" = 7
then u just put

7
------- and that's ur answer
7x-2

Makro · Jun 8, 2009

Kurt, for the last one, I'm not getting it, where does the 1/2 come from?

Makro · Jun 8, 2009

Another one.

Find the integral of (x² + 3/x) from 2 to 1.

I ended up getting 2/1/3, which is half the answer. The whole answer is 2/1/3 + 3ln2.

Where on earth does ln come into the equation?

While we're at it, how would you do $\int_{1}^{e^{3}}\frac{4}{x} dx$

Drongoski · Jun 8, 2009

Makro · Jun 8, 2009

Cheers, first one anyone?

Drongoski · Jun 8, 2009

kurt.physics · Jun 8, 2009

Makro said:
Kurt, for the last one, I'm not getting it, where does the 1/2 come from?

Well, we are trying to get the function in the form f'(x) e^f(x).

Notice that the derivative of the exponent ie x^2 is 2x, by we have only x out the front, so if we write it like 1/2 x 2x, which is the same as x, then we can simply do the integration, then half it.

Makro · Jun 8, 2009

So if I understand correctly, you need to differentiate it to then integrate it? 1/2 x 2x = x which is the original value, but it lets us integrate when its in that form?

$\int \frac{1}{2} \cdot 2xe^{x^2} dx = \frac{1}{2} \int 2xe^{x^2} dx$

$= \frac{1}{2} \cdot e^{x^2} + C$

$= \frac{1}{2} e^{x^2} + C$

Right so I don't understand $\frac{1}{2} \int 2xe^{x^2} dx$
to
$= \frac{1}{2} \cdot e^{x^2} + C$

Makro · Jun 8, 2009

If they should cancel out, why does the 2 disappear and nothing happens with the 1/2?

AND

$\int_{0}^{1}xe^{x^{2}}$

There's a half in the working out, dunno why that's there either. Somebody explain?

Fortify · Jun 8, 2009

The rule is int f'(x)*e^f(x). f'(x) = 2x

So we want the equation to be 2x*e^(x^2)

int x*e^(x^2) is the same as <meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CSAMSON%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w

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<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CSAMSON%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w

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Answer will be 1/2 e^(x^2)

Makro · Jun 8, 2009

Fortify said:
The rule is int f'(x)*e^f(x). f'(x) = 2x

I've never been told this. * means multiply doesn't it? So does .?

so f'(x) TIMES e^f(x) TIMES f'(x).

I must be missing something here.

Fortify · Jun 8, 2009

* is multiply on the computer.
. sometimes may be commonly mistaken as a decimal point when typing on computer.

Fortify · Jun 8, 2009

Makro said:
I've never been told this. * means multiply doesn't it? So does .?

so f'(x) TIMES e^f(x) TIMES f'(x).

I must be missing something here.

OH NO !

I meant f'(x) TIMES e^f(x) FULLSTOP f'(x) is 2x.

Sorry that's my fault I didn't start a new line.

zazzy1234 · Jun 9, 2009

neeeeeeeed some help please

calculate the area bounded by the curve y=2cos2x and the line x=0 and
x= 3pi/4 (diagram if possible as well?)

than yoooooooooooou very much

Fortify · Jun 9, 2009

Here's the diagram:

zazzy1234 · Jun 9, 2009

Fortify said:
Here's the diagram:

thanx 4 the diagram, no all i need is the solution

Fortify · Jun 9, 2009

Yep, hold on. I'm scanning it.

zazzy1234 · Jun 9, 2009

Fortify said:
Yep, hold on. I'm scanning it.

thank uuuuuuuuuu soooooooooo much =)

Logs (differentiation + integration) (1 Viewer)

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