All the points are now combined to create a single graph. We can start looking for patterns by observing the values for the graph of I(b) over the interval
0< b < 5 (approximate values)
0 < b < 1.6 & 3.2 < b < 4.8 the graph is increasing
1.6 < b < 3.2 & 4.8 < b < 5 the graph is decreasing
0.8 < b < 1.6 & 2.4 < b < 3.2 & 4 < b < 4.8 the graph is increasing at a decreasing rate
0 < b < 0.8 & 1.6 < b < 2.4 & 3.2 < b < 4 the graph is increasing at an increasing rate
0 < b < 0.8 & 2.4 < b < 4.8 the graph is concave up
0.8 < b < 2.4 & 4.8 < b < 5 the graph is concave down
Using this information, we can develop a formula
without an integral for I(b). Trials:
f(x) = 3cos(2x) (red)
f(x) = cos(x) (green)
f(x) = 3sin(2x) (blue) (original function)
f(x) = π/2cos(2x+3) + π/2 (red)
f(x) = 3cos(x+π/2)2 (dark blue)
f(x) = 3sin(x)2 (dark blue)
f(x) = -3/2cos(2x)+1.5 (dark blue)
f(x) = 3sin(2x) (green) (original function)
* Compare the values from the
three dark blue functions to those of the original
function. They appear to be equivalent. (Answers are
rounded to 3 significant figures)
Now conduct a similar investigation using the same definite integral
I(b) = but now manipulate the A values.
In this case we will keep the value of A constant and
change the values of b and repeat this several times
for different values of A.
To get the values to plot use the graphing calculator:
FnInt (Y1, X, A, B)
Ex. FnInt (3sin(2x), x, -1, B)
The B values stay the same but the Y
Values will change because we manipulated
Now we must try to generate a formula for in terms of A and B. In the first steps we only made a formula in terms of b. Now we are taking into account both variables to try and reach a more general solution.
Trials:
Start by trying to substitute x for A and B.
Ex. Where A=1 and B=π/3
-3/2cos(2x)+1.5 is our general formula right now
-3/2cos(2π/3)+ 1.5 = 2.25
-3/2cos(21)+ 1.5 = 2.12
(We are given this information from the table above)
Ex. Where A=1 and B= π/2
-3/2cos(2π/2)+ 1.5= 3
-3/2cos(21)+ 1.5= 2.12
(We are given this information from the table above)
From these values we can try subtracting A-B to see if the value will be the same as the value from the definite integral I(b)
2.12-2.25=-1.3
2.12-3= -0.88
These values are similar to the actual value but are the opposite sign. If we subtract B-A instead of A-B, we will get the correct value.
2.25-2.12= 1.31.26
3-2.12= 0.880.876
If we know that by subtracting the values of the two functions from one another then the formula for in terms of a and b is 3cos(B+ π/2)2 – 3cos(A+ π/2)2
Further Examples:
= 3cos(π/3+ π/2)2 – 3cos(2+ π/2)2= -.230
= 3cos(π/6+ π/2)2 – 3cos(3+ π/2)2= .690
= 3cos(π/12+ π/2)2 – 3cos(π/2+ π/2)2= -2.80
In order to reach a more generic function we need to investigate more than one function. For a second function we will use
Start by graphing the function without the integral.
To get the value of the area (definite integral)
Plug FnInt (Y1, X, A, B) into the graphing calculator:
Ex. FnInt (5-4x, x, 0, π/2)= 2.92
Ex. FnInt (5-4x, x, 0, π/6)= 2.07
Observe the values for the graph of I(b) over the
interval 0< b < 5 (approximate values)
0 < b < 1.25 The graph is increasing
1.25 < b < 5 The graph is decreasing
0 < b < 1.25 The graph is increasing at a decreasing rate
1.25 < b < 5 The graph is decreasing at al increasing rate
0< b < 5 The graph is concave down
Now we must try to develop a formula that is equivalent to the definite integral but is integral-less.
Trials:
We know the derivative must be a negative parabola because it is concave down and because it forms a parabola there has to be a squared power in the function.
5-4x (blue)
5-4x2(green)
-4x2(turquoise)
-4x2+3(black)
-4/2x2+3(red)
The equation 5b-2b2 is the best function. This function is the anti-derivative of 5-4x so we know that this is the solution. The function 5b-2b2 holds for all values of B which also confirms that function is indeed correct.
Ex. 5(-1)- 2(-1) 2= -7
Ex. 5(π/2)- 2(π/2) 2= 2.92
Ex. 5(2.25)- 2(2.25) 2= 1.13
Now we can use different values of A to try and find a more general formula:
With our previous knowledge from the
Previous function, we can predict what
The new (more generic) function may
Look like.
Given = 5x-4x2
Let (5b-4b2)- (5a-5a2) be the function
Plug in some examples to see if the same formula applies to this function.
Ex. = ((51)+ 4(12))- ((51)+ 4(12))=0
Ex. = ((51)+ 4(12))- ((5π/2)+ 4(π/22))= 0.081
Ex. = ((5π/2)+ 4(π/22))- ((5-1)+ 4(-12))= 9.92
From this information, we can formulate a shortcut to find for any generic function f(x).
= Whereis the symbol for the anti-derivative
In order to ensure that this shortcut is correct, we must test it out with other example functions. We can try other functions that would require us to apply some of the anti-differentiation rules such as the anti-power rule =
The anti-derivative of this function= 1/3x3
= ((1/3)(13)- ((1/3(03)= .333-0= .333
= ((1/3)(23)-((1/3)(13)= 2.67-.333= 2.33
=
=
=
Now substitute X for values of A and B
Apply the general formula where =
-
A=1 and B=2
-
- = = .25
Now plug in on the calculator:
FnInt((x3-2)/(x3), x, 1, 2
this calculation equals .25 so the generic function does work for this equation.
Assuming that the function of the definite integral itself can be anti-differentiated, then the generic formula we found = should work for all functions at any values of A and B. This helps us to establish a universal rule that can be applied to several different functions. Through this investigation, we proved that you could find a pattern for a mathematical concept by investigating a functions properties and applying our knowledge of calculus and other mathematical concepts to reach an answer. We applied what we already knew about definite integrals, derivatives, anti-derivatives, etc. to develop a formula that now works for all generic functions when you are trying to derive the anti-derivative that is equivalent to the value of the definite integral.