F(3)=1/4 integration

1. Apr 27, 2005

UrbanXrisis

f(3)=1/4

$$\int\frac{dy}{y}=\int (6-2x)dx$$
$$-\frac{1}{y}=6x-x^2+C$$
$$-4=18-9+C$$
$$C=-13$$
$$-\frac{1}{y}=6x-x^2-13$$
$$y=-\frac{1}{6x-x^2-13}$$
$$y=\frac{1}{6x-x^2+13}$$

why?

2. Apr 27, 2005

James R

Looks like the answer is wrong.

3. Apr 27, 2005

UrbanXrisis

I think that the negative was distributed before the Constant was added.. I just dont know why?

4. Apr 27, 2005

whozum

$$\int \frac{dy}{y}$$ is not $$\frac{-1}{y}$$

5. Apr 27, 2005

UrbanXrisis

sorry, it's y^2

f(3)=1/4

$$\int\frac{dy}{y^2}=\int (6-2x)dx$$
$$-\frac{1}{y}=6x-x^2+C$$
$$-4=18-9+C$$
$$C=-13$$
$$-\frac{1}{y}=6x-x^2-13$$
$$y=-\frac{1}{6x-x^2-13}$$
$$y=\frac{1}{6x-x^2+13}$$

why?

6. Apr 27, 2005

whozum

$$-\frac{1}{y}=6x-x^2+C$$

$$y = -\frac{1}{6x-x^2+c}$$

$$-\frac{1}{4} = \frac{1}{9+C}$$

C = 13

7. Apr 27, 2005

UrbanXrisis

sub that back in... you get the same equation as I do...

8. Apr 27, 2005

whozum

I'm sorry, C is -13, and yeah, the one you got is correct.

9. Apr 27, 2005

UrbanXrisis

I dont see how they got that answer...

Last edited by a moderator: May 2, 2017
10. Apr 27, 2005

whozum

You wrote the answer in correct in your first post, and unless theres a new math system where

C + 9 = 4, and C = 18 theyre wrong. C is -13

11. Apr 27, 2005

UrbanXrisis

I took the answers right off of the college-board website! wow...

12. Apr 27, 2005

Hippo

Nobody's infall-yable.