# Integral

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• Feb 10th 2010, 12:45 AM
Awsom Guy
Integral
A curve has its gradient functions as (2ax-b). It passes through the origin and (2,8) is its maximum point. Find a, b, and the equation of the curve.

I know a couple of things that its got something to do with the word maximum and make sure you understand and (2,8) as its maximum point.
I have no working to give as I have no idea where to begin.
Thanks
• Feb 10th 2010, 12:49 AM
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Quote:

Originally Posted by Awsom Guy
A curve has its gradient functions as (2ax-b). It passes through the origin and (2,8) is its maximum point. Find a, b, and the equation of the curve.

I know a couple of things that its got something to do with the word maximum and make sure you understand and (2,8) as its maximum point.
I have no working to give as I have no idea where to begin.
Thanks

If $f'(x) = 2ax - b$ and $(2, 8)$ is the maximum, then $f'(2) = 0$.

Therefore $2a(2) - b = 0$

$b = 4a$.

$f(x) = \int{2ax - b\,dx}$

$= ax^2 - bx + c$

And since the curve passes through $(0, 0)$ and $(2, 8)$ substitute these values into $f(x)$ and you will end up with some equations you can solve simultaneously for $a, b, c$.
• Feb 10th 2010, 12:54 AM
Awsom Guy
Sorry I dont get that and whats the big F looking thing sorry that stuff I haven;t learnt. :)
• Feb 10th 2010, 01:13 AM
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Quote:

Originally Posted by Awsom Guy
Sorry I dont get that and whats the big F looking thing sorry that stuff I haven;t learnt. :)

I haven't used a big F...
• Feb 10th 2010, 01:15 AM
Awsom Guy
umm.. the big \int thing.
• Feb 10th 2010, 01:23 AM
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Quote:

Originally Posted by Awsom Guy
umm.. the big \int thing.

It's the integral sign, and it stands for the indefinite integral, also known as the antiderivative.

So I'm saying take the antiderivative of that function.
• Feb 10th 2010, 01:26 AM
Awsom Guy
ok thanks
• Feb 10th 2010, 01:27 AM
Awsom Guy
but where does the dx come from.
• Feb 10th 2010, 01:29 AM
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Quote:

Originally Posted by Awsom Guy
but where does the dx come from.

It's just notation.

It means take the antiderivative of the function with respect to the variable $x$.

It's important for when you learn the reverse chain rule.
• Feb 10th 2010, 01:30 AM
Awsom Guy
so I get:
C=0 and:
8=4a^2-2b+C
How can I simultanously solve this.
• Feb 10th 2010, 01:47 AM
Prove It
Quote:

Originally Posted by Awsom Guy
so I get:
C=0 and:
8=4a^2-2b+C
How can I simultanously solve this.

You know that $c = 0$ and $b = 4a$. Substitute them in.
• Feb 10th 2010, 01:49 AM
Awsom Guy
does a equal to: (8)/(4a-8)
thanks
• Feb 10th 2010, 01:50 AM
Awsom Guy
these are meant to be the answers:
a=-2
b=-8
y=-(2x^2)-8x
• Feb 10th 2010, 01:55 AM
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Quote:

Originally Posted by Awsom Guy
these are meant to be the answers:
a=-2
b=-8
y=-(2x^2)-8x

You should be getting a quadratic equation in $a$. You'd need to factorise or use the Quadratic formula to evaluate $a$.
• Feb 10th 2010, 01:57 AM
Awsom Guy
Sorry I am just not getting this, I'll just skip this question thanks anyways.
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