 Level: International Baccalaureate
 Subject: Maths
 Word count: 2337
IB Math IA evaluating definite integrals
Extracts from this document...
Introduction
In this assignment we are being asked to discover a way to evaluate definite integrals independently of our calculator. In order to proceed with the investigation, we must identify the meaning of a definite integral.
A definite integral is defined as the area between a curve in the given xaxis in a given interval. The definite integral of the function f from x = a to x = b gives a way to find the product of (ba) and f(x), even if f is not a constant. The definite integral is also defined as an average rate of change and can be written as:
Given the information above, investigate the definite integral of 3sin(2x) defined as:
I(b) =
← Graph of 3sin(2x)
We can begin the investigation by plotting points on a graph to represent values of the definite integral. For this trial we will keep A=0 and B will be manipulated.
Start by plugging in the equation to the calculator:
Ex. 3sin(2x) where x= π/2
3sin(2π/2)= 0
Ex. 3sin(2x) where x= π/6
3sin(2π/6)= 2.60
These are the y values for the function. Then graph
the (x,y) coordinates. The area above the x=axis and
in between the two points is considered the definite integral
or area under the curve. Next, shade the area
under the curve from 0
Middle
1.5
π/6
1.37
1.73
0.69
2.25
π/12
1.92
2.28
0.141
2.80
1
0
0.356
2.06
0.876
2
0.356
0
2.421
0.520
3
2.06
2.42
0
2.94
4
0.406
0.762
1.66
1.28
1
0
0.356
2.06
0.876
2
0.356
0
2.42
0.520
3
2.06
2.42
0
2.94
4
0.406
0.762
1.66
1.28
2.25
0.308
0.664
1.76
1.18
2.5
1.05
1.41
1.01
1.93
2.75
1.69
2.04
0.377
2.56
3.25
2.09
2.45
0.025
2.96
3.5
1.76
2.11
0.309
2.63
2.25
0.308
0.664
1.76
1.18
2.5
1.05
1.41
1.01
1.93
2.75
1.69
2.04
0.377
2.56
3.25
2.09
2.45
0.024
2.96
3.5
1.76
2.11
0.309
2.63
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 AB to see if the value will be the same as the value from the definite integral I(b)
2.122.25=1.3
2.123= 0.88
These values are similar to the actual value but are the opposite sign. If we subtract BA instead of AB, we will get the correct value.
2.252.12= 1.31.26
32.12= 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 (54x, x, 0, π/2)= 2.92
Ex. FnInt (54x, x, 0, π/6)= 2.07
B  Y (A=0) 
0  0 
π/2  2.92 
π/3  3.04 
π/4  2.69 
π/6  2.07 
π/12  1.17 
π/2  12.8 
π/3  7.43 
π/4  5.02 
π/6  3.17 
π/12  1.45 
1  3 
2  2 
3  3 
4  12 
1  7 
2  18 
3  33 
4  52 
2.25  1.13 
2.5  0 
2.75  1.38 
3.25  4.88 
3.5  7 
2.25  21.4 
2.5  25 
2.75  28.9 
3.25  37.4 
3.5  42 
Conclusion
Ex. = ((5π/2)+ 4(π/22)) ((51)+ 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 antiderivative
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 antidifferentiation rules such as the antipower rule =
The antiderivative of this function= 1/3x3
= ((1/3)(13) ((1/3(03)= .3330= .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((x32)/(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 antidifferentiated, 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, antiderivatives, etc. to develop a formula that now works for all generic functions when you are trying to derive the antiderivative that is equivalent to the value of the definite integral.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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