Integration (1 Viewer)

yeyjasminetime · Oct 4, 2013

Hi,

Find y as a function of x, if y' = 1/4x and y= 1 and y = 1 when x = e2. What is the x intercept of this curve?

If someone could explain this to me, it would be great. Thank you.

HeroicPandas · Oct 4, 2013

$y' = \frac{dy}{dx } \\ \frac{dy}{dx} = \frac{1}{4x} = \frac{1}{4}\cdot \frac{1}{x}$
$dy = \frac{1}{4}\cdot \frac{1}{x}dx$

$\int dy = \int \frac{1}{4}\cdot \frac{1}{x}dx$

$\int dy = \frac{1}{4}\int \frac{1}{x}dx$

$\therefore y = \frac{1}{4}lnx + C$

When y = 1, x = e2 (is this e squared? im gonna assume its e squared)

$\therefore1 = \frac{1}{4}ln(e^2) +C \rightarrow \therefore C = \frac{1}{2}$

Therefore y = .......

What is an x-intercept? It is when the curve cuts the x-axis, so the y-coordinate must be...?

hit patel · Oct 4, 2013

^^^Yep

yeyjasminetime · Oct 4, 2013

Yes! That's where I got up to. The x intercept is when y = 0. I'm just confused about what C is and what it's got to do with anything?

Thanks heroic panda

HeroicPandas · Oct 4, 2013

yeyjasminetime said:
Yes! That's where I got up to. The x intercept is when y = 0. I'm just confused about what C is and what it's got to do with anything?

Thanks heroic panda

cool!

yeyjasminetime · Oct 4, 2013

So what does the C represent?

HeroicPandas · Oct 4, 2013

the reason why u have 'C' is that the original function could have a constant added to it

eg. y = x+1, y' = 1
If u integrate back, u get:
y = x + C

eg. y = x, y' = 1
If u integrate back, u get:
y = x + C

C is determined by a relation between x and y

Another example, y' = x^2 /2
Integrate with respect to x
y = x + C
U dont know if the original function could have been y = x+2 or y = x+ 223 or y = x + 2013, etc

THats why the C is there

yeyjasminetime · Oct 4, 2013

Comprende! Thank you

HeroicPandas · Oct 4, 2013

yeyjasminetime said:
Comprende! Thank you

No problem!

yeyjasminetime · Oct 4, 2013

So how do you end up with x=e^-2

HeroicPandas · Oct 4, 2013

$y = \frac{1}{4}lnx + \frac{1}{2}$

x-intercept occurs when y = 0,

So we let y = 0,

$\therefore 0 = \frac{1}{4}lnx + \frac{1}{2}$

Make 'lnx' the subject and put it on the left so its easier to see,

$lnx = -2$

$\therefore x = e^{-2}$ (THIS PART, changing logs to exponential, DO U UNDERSTAND?)

yeyjasminetime · Oct 4, 2013

THATS EXACTLY THE PART I DONT UNDERSTAND! I think I should revise my log and Ln. My teacher said to use Ln instead of log.

:/

yeyjasminetime · Oct 4, 2013

Oh wait! I get it.

yeyjasminetime · Oct 4, 2013

BUT WHERE DOES e COME FROM? ):

Menomaths · Oct 4, 2013

$ln = log_e$

$log_{a}b = x$

$a^x = b$

$Similarly:$

$log_{e}x = -2$

$e^{-2} = x$

HeroicPandas · Oct 4, 2013

http://community.boredofstudies.org/showthread.php?t=311248

Read post #13, if you will do not understand

Integration (1 Viewer)

yeyjasminetime

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HeroicPandas

Heroic!

hit patel

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