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Math Crew: Help!
- Thread starter abhigames
- Start date
area under a curve is defined as the integral of the function. therefore if you integrate the function over the your limits, you should get the area under that curve. From there find the area under the curve you are trying to look for by subtracting it from the total area. Once you have that go backwards and take the derivative of that area to get the function of the curve.
MATH TEAM PRIDE. lol
Okay, here's the information from your pdf file: (I changed variable a to variable x)
f(a) = Top Curve = -x + 1
f(b) = Bottom Curve
Integral of f(a) = 0.5 (It's a triangle with intercepts at (0,1) and (1,0). Area = 1/2 (b) * (h) = 1/2
GIVENS
A = area between f(a) and f(b) = 0.2335
B = area between f(b) and axis = 0.2665
(A + B should = 0.5 and they do)
Now, f(a) is a straight line. f(b) has intercepts (0,1) and (1,0) and HAS to be at least polynomial. F(b) cannot be straight line and there should be a polynomial solvable equation. Anything more should need more information, so it's logical to assume that polynomial is answer.
Therefore, we can assume that it, f(b) = ax^2 + bx + c with solutions at least at (0,1) and (1,0)
f(b) = ax^2 + bx + c WITH inserted x
1 = c (x = 0)
0 = a + b + c (x = 1)
Then, a + b = -1 and a = - b - 1 OR b = - a - 1
Obviously, we need MORE information. Well, we know that the integral of f(b) from 0 to 1 equals the area of B which is 0.2665.
f(b) = ax^2 + bx + 1 (we know c = 1 now)
f(b) = ax^2 - ax - x + 1
Integral of f(b) = (a/3)x^3 - (a/2)x^2 - (1/2)x^2 + x = 0.2665
From 0 to 1, (a/3) - (a/2) - (1/2) + 1 = 0.2665
-(a/6) + (1/2) = 0.2665, a = 1.401
b = -2.401
c = 1
Thus, f(b) = 1.401x^2 - 2.401x + 1 = (1.401)b^2 - (2.401)b + 1 (in your units)
Okay, here's the information from your pdf file: (I changed variable a to variable x)
f(a) = Top Curve = -x + 1
f(b) = Bottom Curve
Integral of f(a) = 0.5 (It's a triangle with intercepts at (0,1) and (1,0). Area = 1/2 (b) * (h) = 1/2
GIVENS
A = area between f(a) and f(b) = 0.2335
B = area between f(b) and axis = 0.2665
(A + B should = 0.5 and they do)
Now, f(a) is a straight line. f(b) has intercepts (0,1) and (1,0) and HAS to be at least polynomial. F(b) cannot be straight line and there should be a polynomial solvable equation. Anything more should need more information, so it's logical to assume that polynomial is answer.
Therefore, we can assume that it, f(b) = ax^2 + bx + c with solutions at least at (0,1) and (1,0)
f(b) = ax^2 + bx + c WITH inserted x
1 = c (x = 0)
0 = a + b + c (x = 1)
Then, a + b = -1 and a = - b - 1 OR b = - a - 1
Obviously, we need MORE information. Well, we know that the integral of f(b) from 0 to 1 equals the area of B which is 0.2665.
f(b) = ax^2 + bx + 1 (we know c = 1 now)
f(b) = ax^2 - ax - x + 1
Integral of f(b) = (a/3)x^3 - (a/2)x^2 - (1/2)x^2 + x = 0.2665
From 0 to 1, (a/3) - (a/2) - (1/2) + 1 = 0.2665
-(a/6) + (1/2) = 0.2665, a = 1.401
b = -2.401
c = 1
Thus, f(b) = 1.401x^2 - 2.401x + 1 = (1.401)b^2 - (2.401)b + 1 (in your units)
Navkarn
Member
- Messages
- 658
haha yesssssss math nerds unite!!!
check out my calculus quotient rule freestyle, only true math geeks appreciate and understand it! lol
http://youtube.com/watch?v=DdV2UZV7AoA
low d high minus high d lowwwwwwwwwwww, square the bottom and away we go! ;D
check out my calculus quotient rule freestyle, only true math geeks appreciate and understand it! lol
http://youtube.com/watch?v=DdV2UZV7AoA
low d high minus high d lowwwwwwwwwwww, square the bottom and away we go! ;D
HAHAHAH YES!
But respect Navkarn! Very few could make it sound like a legit song. I remember in Math Team, we made math raps about integrals, mysterious "e", little p (statistical analysis) and z-tests but alas, those were only raps and require no skill to sing.
But respect Navkarn! Very few could make it sound like a legit song. I remember in Math Team, we made math raps about integrals, mysterious "e", little p (statistical analysis) and z-tests but alas, those were only raps and require no skill to sing.