Question

A curve is such that \[{dy \over dx} = {3 \over (1+2x)^2}\] and the point (1, 1/2) lies on the curve.

Find the equation of the curve.

Answer 1

Isn't it you're meant to integrate first?

Answer 2

u integrate it to find equation of thee curve,, Then u substitute the points to find the constant of integration

Answer 3

do u know how to integrate this?

Answer 4

\[dy \div 3 = dx \div (1+2x)^{2}\]
Now integrate both sides.
Put t = (1+2x). Now the integration should be easy.
Find the constant of integration by putting (1,0.5) in the curve.
This should give you the answer.

Answer 5

Hmm never heard of this way... How do you do it?

Answer 6

\[\huge{y=\int\limits{\frac{3}{(1+2x)^3}dx}}\]

Answer 7

If you put t = 1+2x, then dt/2 = dx (by differentiating both sides). Now, the RHS becomes:
1/2 dt / (t^2). Now integrate it. It's called the variable separation method - Separate the variables involved, one variable on either side of the equation and then integrate both sides.

Answer 8

use substitution
let u=1+2x
du=2dx
so the integral becomes
\[\int\limits{\frac{3}{u^3} \times \frac{du}{2}}\]

Answer 9

Hmm Still lost :/

Answer 10

simple integration u just move the dx to the other side so u can integrate with respect to x

Answer 11

I got
\[{ 3(1+2x)^{-1} \over -1} +c\]

Answer 12

But it says that is the first step... and that you have to divide by 2 to equate it to y. How come?

Answer 13

@lalaly

Answer 14

yeah do u know integration by substitution?

Answer 15

No..

Answer 16

ok let u=1+2x
so du=2dx
so dx=du/2

so u substitute thos values to make it easier to integrate\[\huge{\int\limits{\frac{3}{u^3} \times \frac{du}{2}}}\]

Answer 17

now u integrate that with respect to u
then resubstitute the u=1+2x when ur done

Answer 18

How?

Answer 19

i suggest u learn the basic methods of integration,
by substitution, by parts .... etc... i cant do this for u if u dont know it at all

Answer 20

I learned it a different way...

Answer 21

@lalaly

Answer 22

how?

Answer 23

\[\int\limits 3 (1 +2x)^{-2}\]
\[3 \int\limits (1+2x)^{-2}\]
\[3 \int\limits (1+2x)^{-2} \times 1/2\]
\[3/2 (1+2x)^{-1}\]

Answer 24

okthats right, then nevermind :) but remember for an indefinite integral u add a constant
so\[\huge{y=\frac{3}{2}\frac{1}{1+2x)^2}+C}\]

Answer 25

Oh, yeh. But how do I substitute the values?

Answer 26

to the power of -2 not -1 .. in ur last post

Answer 27

But it's integrated.. and it says that's right..?

Answer 28

for the point(1,1/2)

\[\huge{\frac{1}{2}=\frac{3}{2(1+2(1))^2}}+C\]
now just solve for C

Answer 29

Ok, thanks :)

Answer 30

But it is ^-1

Answer 31

no its -2 .. because it was -3 integrating that u just add 1 to the power which becomes -2

Answer 32

But it was originally ^-2

Answer 33

oh latex makes it look like a 3,, im sorry i thought it was 3 never mind lol

Answer 34

That's Ok. Thanks!