[EventHorizon]

As you know, a quadratic is of the form ax^2+bx+c. Given the equation, it's easy to integrate to find the area under the curve between two values. Lets see if you can find the area under the curve without knowing the function.

You know the following three pieces of information (where g, h, j, and k are real numbers):

f(x) is quadratic

f(j) - f(j+2k) = g

f(j+2k) + 2*f(j+k) = h

In terms of g, h, j, and k, what is the value of integrating f(x) between j and j+2k? How did you find it?

(Yes, it is possible to solve this, and, yes, I am being really sadistic)

Edited September 14, 2010 by EventHorizon

[Guest]

Answer is k*(4h+2g)/6

Assuming f(x)=ax^2+bx+c .............(1)

we get value of

g= -4ak^2-4ajk-2bk .............(2)

h= 3aj^2+8ajkj\=6ak^2+3jb+4bk+3c ............(3)

After integrating ax^2+bx+c wrt x, we get

ax^3/3 + bx^2/2 +cx. ...............(4)

Substituting value of j and j=2k, in (4) we get final equation as

(12akj^2+24ajk^2+16ak^3+12bjk+12bk^2+12ck)/6

After simplifying we get

[4*(3aj^2 +8ajk+6ak^2+3jb+4bk+3c)+2(-4ajk-4ak^2-2bk)]*k/6 ....................(5)

substituting values of g and h from (3) and (2) in (5) we get

integral value is k*(4h+2g)/6 ..........Q.E.D.

[EventHorizon]

Answer is k*(4h+2g)/6

Assuming f(x)=ax^2+bx+c .............(1)

we get value of

g= -4ak^2-4ajk-2bk .............(2)

h= 3aj^2+8ajkj\=6ak^2+3jb+4bk+3c ............(3)

After integrating ax^2+bx+c wrt x, we get

ax^3/3 + bx^2/2 +cx. ...............(4)

Substituting value of j and j=2k, in (4) we get final equation as

(12akj^2+24ajk^2+16ak^3+12bjk+12bk^2+12ck)/6

After simplifying we get

[4*(3aj^2 +8ajk+6ak^2+3jb+4bk+3c)+2(-4ajk-4ak^2-2bk)]*k/6 ....................(5)

substituting values of g and h from (3) and (2) in (5) we get

integral value is k*(4h+2g)/6 ..........Q.E.D.

Correct. I was hoping it would last longer, but looking at your work I see that it was simpler than I thought.

I tried to add in a little misdirection (which only simplified the problem), but the main point of this is to show that you can find the exact integral for an area under a quadratic if you know the value of the function at both ends of the area and the middle.

Integral of a quadratic f(x) between j and j+2 = ( f(j) + 4*f(j+1) + f(j+2) )/3

This pattern of 1,4,1 is used as a quadratic approximate for areas under curves for data sampled at constant intervals.

Since this was found so simply, I'll post another named "Integrate an unknown polynomial"

Edited September 15, 2010 by EventHorizon

[Guest]

Correct. I was hoping it would last longer, but looking at your work I see that it was simpler than I thought.

I tried to add in a little misdirection (which only simplified the problem), but the main point of this is to show that you can find the exact integral for an area under a quadratic if you know the value of the function at both ends of the area and the middle.

Integral of a quadratic f(x) between j and j+2 = ( f(j) + 4*f(j+1) + f(j+2) )/3

This pattern of 1,4,1 is used as a quadratic approximate for areas under curves for data sampled at constant intervals.

Since this was found so simply, I'll post another named "Integrate an unknown polynomial"

Hi I know that this might sound rather stupid but isn't the integral of a quadratic between a and b F(b) -F(a)? In this case wont it become f(j+2k)-f(j)or -g? Its been a long time since I have done Maths (at least 13 years) so It would be great if you could explain it in simple terms :-).

[EventHorizon]

Hi I know that this might sound rather stupid but isn't the integral of a quadratic between a and b F(b) -F(a)? In this case wont it become f(j+2k)-f(j)or -g? Its been a long time since I have done Maths (at least 13 years) so It would be great if you could explain it in simple terms :-).

You can evaluate a definite integral for a quadratic without performing the integration. You simply need to evaluate the function at three specific places and plug them into an equation. Which means as long as you can sample from the quadratic function, you don't need to know the function to integrate it (for definite integrals at least).

Here's a quick proof that F(x+2)-F(x) = (f(x)+4*f(x+1)+f(x+2))/3 for any value of x. (which can be generalized for j to j+2k):

f(x) = ax^2 + bx + c

F(x) = ax^3/3 + bx^2/2 + cx

F(x+2)-F(x) = a(x+2)^3/3 + b(x+2)^2/2 + c(x+2) - ax^3/3 - bx^2/2 -cx

F(x+2)-F(x) = (6ax^2 +12ax + 8a)/3 + (4bx + 4b)/2 + 2c

F(x+2)-F(x) = (6ax^2 +12ax + 8a)/3 + 2bx + 2b + 2c

F(x+2)-F(x) = (6ax^2 +12ax + 8a + 6bx + 6b + 6c)/3

f(x) = ax^2 + bx + c

f(x+1) = a(x+1)^2 + b(x+1) + c = ax^2 + 2ax + a + bx + b + c

f(x+2) = a(x+2)^2 + b(x+2) + c = ax^2 + 4ax + 4a + bx + 2b + c

f(x)+4*f(x+1)+f(x+2) = ax^2 + bx + c + 4ax^2 + 8ax + 4a + 4bx + 4b + 4c + ax^2 + 4ax + 4a + bx + 2b + c

f(x)+4*f(x+1)+f(x+2) = 6ax^2 + 12ax + 8a + 6bx + 6b + 6c

F(x+2)-F(x) = (6ax^2 +12ax + 8a + 6bx + 6b + 6c)/3 = (f(x)+4*f(x+1)+f(x+2))/3

[Guest]

Thanks! that was interesting :-)