Question

I just want to check my answer..
The gradient of a curve is given by dy/dx = kx^2+6, where k is a constant, given that the curve passes through (1,6) and (-1,2).
Find the value of y in terms of x.

I have found k= 15
so it's y= 15x^3/3 + 6x +c
or y= 5x^3 + 6x + c .

Answer 1

by integration.

Answer 2

i got k=-12 and c=4.
Check your work again.
I know mine's right cuz you can easily verify it by plugging the values into the equation.

Answer 3

oh okay , imma check it

Answer 4

sure

Answer 5

wrong answer :/
I did like this :
integral (kx^2+6) dx
y = kx^2/3 + 6x + c
6 = k(1)^2/3 + 6 (1) + c ---> (1)
2 = k(-1)^2/3 + 6 (-1) + c ---> (2)
by subtracting (1) - (2)
2 = 2/3k + 12
2-12 = 2/3k
k= -15 .. ?

Answer 6

and c=3

Answer 7

what's the integral of x^3?

Answer 8

x^4/4 ?

Answer 9

sorry. the integral of x^2?

Answer 10

x^3/3 .. I just noticed that^ But I calculated it x^3/3

Answer 11

6-2?

Answer 12

4

Answer 13

yeah but that's the mistake. you've used 2 in your calculation...

Answer 14

oh pellet , silly mistakes can flutter the whole calculation :s

Answer 15

yeah lol

Answer 16

sorry I didn't concentrate, thanks :D

Answer 17

word filters - you gotta hate them ;)

Answer 18

haha yeah F U C K IT ;)

Answer 19

lol