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# integration help(i think?) Tweet

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1. integration help(i think?)
1. The gradient of a curve is given by dy/dx=3x-4. Find the equation of the curve given that it passes through the point (3,1)

so integrate 3x-4 so= 3x^2/2-4x. can i get rid of the 2 on the denominator?
Last edited by dongonaeatu; 12-04-2012 at 17:35.
2. Re: integration help(i think?)
(Original post by dongonaeatu)
1. The gradient of a curve is given by dy/dx=3x-4. Find the equation of the curve given that it passes through the point (3,1)

so integrate 3x-4 so= 3x^3/3-4x. can i get rid of the 3 on the denominator?
Where are you getting the 3 from?
3. Re: integration help(i think?)
To integrate you increase the power by one and divide by the new power. So yes you can cancel the 3 and would therefore have y=x^3-4x+d

Don't forget to include the constant!!!!

Solve using the values you know to find d.
4. Re: integration help(i think?)
Is your original expression supposed to be dy/dx = 3x - 4, i.e. the equationn of a straight line? If you've integrated incorrectly.

If it is supposed to be dy/dx = 3x^2 - 4, i.e. a quadratic, then yes, this gives you y = (3x^3)/3 - 4x + c. Of course you can cancel the 3s in the first term - it is just three times something divided by three.
5. Re: integration help(i think?)
Isn't it divided by two, not three :P

You divide by the new power bro

Also, +k on the end.
6. Re: integration help(i think?)
(Original post by CD315)
Where are you getting the 3 from?
sorry, i meant 2.

is the final answer for the equation of the curve y=3x^2/2-4x-1/2
Last edited by dongonaeatu; 12-04-2012 at 17:28.
7. Re: integration help(i think?)
(Original post by tamimi)
Isn't it divided by two, not three :P

You divide by the new power bro

Also, +k on the end.
is the final answer for the equation of the curve y=3x^2/2-4x-1/2
8. Re: integration help(i think?)
That's what I make it.
9. Re: integration help(i think?)
(Original post by dongonaeatu)
is the final answer for the equation of the curve y=3x^2/2-4x-1/2
I believe it is Either that or we're both wrong.
10. Re: integration help(i think?)
Yep I seem to get -1/2. You don't need to get rid of the two, just sub in your x and y values after you've integrated. Best way just in case you make a mistake in getting rid of the 2.
11. Re: integration help(i think?)
(Original post by mr tim)
Yep I seem to get -1/2. You don't need to get rid of the two, just sub in your x and y values after you've integrated. Best way just in case you make a mistake in getting rid of the 2.
thanks tim, are you good at trigonometry?
12. Re: integration help(i think?)
(Original post by dongonaeatu)
thanks tim, are you good at trigonometry?
if you can tell me the question then I'll or someone else will tell you how to answer it.
13. Re: integration help(i think?)
(Original post by mr tim)
if you can tell me the question then I'll or someone else will tell you how to answer it.
http://www.thestudentroom.co.uk/show...260&p=37114590