Anti-differentiation

• February 9th 2010, 11:35 PM
Awsom Guy
Anti-differentiation
I can't do this questio although my teacher helped me I still can't do it, please check.

y''= x+2 and if x=3, y'=7 and y=12.
(hint: evaluate constants as they are created)

My failed attempt:

y'=(x^2)/(2) + 2x + C = 7 when x=3.
9/2 + 6 + C=7
C=-3.5

And I also know that (x^3)/(6) +x^2 +7 = 12
That is the second derivative.
I hope what I have written makes sense.
Thanks
• February 9th 2010, 11:43 PM
dedust
Quote:

Originally Posted by Awsom Guy
I can't do this questio although my teacher helped me I still can't do it, please check.

y''= x+2 and if x=3, y'=7 and y=12.
(hint: evaluate constants as they are created)

My failed attempt:

y'=(x^2)/(2) + 2x + C = 7 when x=3.
9/2 + 6 + C=7
C=-3.5

And I also know that (x^3)/(6) +x^2 +7 = 12
That is the second derivative.
I hope what I have written makes sense.
Thanks

we have $y' = (x^2)/(2) + 2x - 3.5$

then

$y = (x^3)/(6) +x^2 - 3.5x + K$

when $x = 3$

$12 = (3^3)/(6) +3^2 - 3.5(3) + K$
• February 9th 2010, 11:46 PM
Awsom Guy
Quote:

Originally Posted by dedust
we have $y' = (x^2)/(2) + 2x - 3.5$

then

$y = (x^3)/(6) +x^2 - 3.5x + K$

when $x = 3$

$12 = (3^3)/(6) +3^2 - 3.5(3) + K$

Wait where did the K come from, sorry I dont follow.
• February 10th 2010, 12:13 AM
Awsom Guy
I am soooo sorry for being dumb...de dust has given me what I have to do thanks.
• February 10th 2010, 12:14 AM
dedust
Quote:

Originally Posted by Awsom Guy
Wait where did the K come from, sorry I dont follow.

another constant from the indefinite integral.