• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# 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 x-axis 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 (b-a) 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(2 π/2)= 0

Ex. 3sin(2x) where x= π/6

3sin(2 π/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(2 π/3)+ 1.5 = 2.25

-3/2cos(2 1)+ 1.5 = 2.12 (We are given this information from the table above)

Ex. Where A=1 and B= π/2

-3/2cos(2 π/2)+ 1.5= 3

-3/2cos(2 1)+ 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= 1.3 1.26
3-2.12= 0.88 0.876

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

 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 π/2)+ 4(π/22))= 0.081

Ex. = ((5 π/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). = Where is 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.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

ï¿½ï¿½Ó¨ï¿½2:#ï¿½(tm)ï¿½Vï¿½Bï¿½ï¿½Q:"Lï¿½"1ï¿½Iï¿½[0ï¿½0ï¿½%ï¿½bï¿½eï¿½e1/2Âï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½]ï¿½x ï¿½)|N"Y,S|Vï¿½Vï¿½Eï¿½Nï¿½Oï¿½ï¿½ï¿½-Rï¿½ï¿½iï¿½~Õ·jkï¿½ï¿½ï¿½Z'ï¿½ï¿½:Æºï¿½z-ï¿½ï¿½ ï¿½ï¿½F%ï¿½&]ï¿½ï¿½ï¿½^ï¿½ï¿½Yï¿½Zï¿½ï¿½lï¿½mï¿½v)ï¿½iï¿½:X8ï¿½:98;"ï¿½ï¿½ï¿½ï¿½ï¿½"{xxï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½@ï¿½ï¿½5ï¿½ï¿½W ï¿½ ï¿½ b ï¿½ï¿½Sï¿½"1/4 }vYï¿½""ï¿½ï¿½D'ï¿½Pb ï¿½8ï¿½3/4ï¿½&"&F&Y& %oï¿½<Kmï¿½{0ï¿½']-ï¿½ Y[×³ ï¿½ï¿½ï¿½ï¿½ï¿½0ï¿½ï¿½ï¿½ï¿½ï¿½-:|P;ï¿½5o3}Xï¿½Hï¿½Qï¿½cWï¿½oï¿½Kï¿½ï¿½-'/U(S*W(r)P(c)T9(c)Zï¿½yï¿½ï¿½tï¿½(tm)ï¿½ï¿½5,ï¿½ï¿½1/4ï¿½"Χï¿½ï¿½6oï¿½1/4p(r)(c)ï¿½ï¿½ï¿½ï¿½ï¿½-ï¿½Kï¿½ï¿½\ï¿½ï¿½6ï¿½&ï¿½-ï¿½Qï¿½ï¿½ï¿½ï¿½ì­nï¿½[ =.1/2...}sï¿½ï¿½ï¿½w-ï¿½};ï¿½xï¿½@dï¿½ï¿½pì£°ï¿½ï¿½jï¿½1/4c4ckOï¿½ï¿½>ï¿½ï¿½ï¿½kï¿½ï¿½ï¿½[/z'"_ï¿½1/4:ï¿½:ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½'wï¿½ï¿½ï¿½+gï¿½}ï¿½?ï¿½<>ï¿½ï¿½ï¿½ï¿½"ï¿½yï¿½ï¿½ï¿½;_jï¿½ï¿½] Yï¿½]Vï¿½*ï¿½Mbï¿½ï¿½{ï¿½ï¿½ï¿½ï¿½o686Ý¶jï¿½ï¿½wï¿½'%:ï¿½ï¿½>@bP\$ï¿½sï¿½ï¿½ï¿½2* ï¿½}#ï¿½ï¿½ï¿½ï¿½dq_ï¿½}"*ï¿½4ï¿½?ï¿½ï¿½'Îï¿½-!ï¿½1ï¿½\ï¿½4ï¿½ï¿½Uï¿½Í1/2ï¿½ï¿½ï¿½Ûï¿½0ï¿½0?Iï¿½^ï¿½ÐQï¿½D,A1/4Nï¿½ï¿½7iqYe9 y %E %^eHï¿½ï¿½ ï¿½[ jtj6hUhï¿½ï¿½ï¿½ï¿½ï¿½Ö·504T3'56ï¿½4e0Úï¿½ï¿½/Zï¿½XNX Ywï¿½\ï¿½ï¿½ï¿½Ud-cï¿½@qï¿½sï¿½u-rawŸ.ï¿½1/2tï¿½ï¿½gï¿½Wï¿½-/oi ï¿½ï¿½+ï¿½^ï¿½Úï¿½ï¿½ï¿½A6ï¿½rTFï¿½×ï¿½ï¿½maeï¿½ï¿½n'ï¿½Qï¿½ï¿½[1Eï¿½ï¿½q*ï¿½ï¿½"sï¿½ï¿½I&ï¿½ï¿½ï¿½ï¿½)WR3ï¿½Ú¦qï¿½}Lï¿½ï¿½8''ï¿½mï¿½xï¿½ï¿½~ï¿½\ï¿½R Eï¿½ qï¿½"ï¿½ ï¿½l|?ï¿½tlï¿½ï¿½Z(r)ï¿½ï¿½"xï¿½Z(c)q(tm)]ï¿½ï¿½ Ê"iU Nï¿½>yï¿½Ùï¿½Ñï¿½suï¿½ï¿½ï¿½ï¿½[^4n5 4ï¿½\ n9r(c)ï¿½ï¿½-"*ï¿½ï¿½>n'uï¿½tR"Ê¯?ï¿½ï¿½Ý­r+ï¿½ï¿½ï¿½ï¿½ï¿½mï¿½~ï¿½ï¿½"ï¿½ï¿½ï¿½ï¿½æ"-z g?jy6ï¿½S~ï¿½4g1/4ï¿½Yï¿½Ä4"r/ï¿½^%ï¿½>ï¿½ï¿½ï¿½ï¿½ï¿½;...ï¿½Wg-gï¿½ï¿½ï¿½k>3/4Yï¿½ï¿½ï¿½ï¿½_ï¿½K _eVï¿½"4kï¿½ï¿½O×·ï¿½Ûï¿½i`ï¿½Ô¦ï¿½ï¿½>"...ï¿½ ï¿½ï¿½4ï¿½ï¿½E'Pï¿½ï¿½8t!FÓï¿½ï¿½'qï¿½ï¿½ ï¿½4|4kï¿½'ï¿½ï¿½stï¿½ï¿½... G ï¿½Lï¿½ï¿½XY_ï¿½ï¿½qï¿½9eï¿½Lï¿½)<{yKï¿½.ï¿½ <ï¿½ï¿½,1/4,ï¿½ï¿½dMï¿½}ï¿½\$ï¿½IyHï¿½ï¿½@2ï¿½ï¿½ï¿½rï¿½ï¿½- ï¿½ "ï¿½ï¿½Juï¿½ï¿½*Îªï¿½jï¿½ï¿½"zï¿½Fï¿½ï¿½ï¿½-ï¿½Ö¦ï¿½ï¿½Nï¿½nï¿½ï¿½ï¿½>ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½9ï¿½tcYSï¿½ï¿½Kï¿½kï¿½,|,Uï¿½ï¿½VSÖ­6yï¿½^Hï¿½ï¿½ï¿½=ï¿½orï¿½tttqï¿½ï¿½<ï¿½Rï¿½ï¿½ï¿½Ntï¿½ï¿½;ï¿½ï¿½Üï¿½ï¿½ï¿½ï¿½>V6Ê´ï¿½Eï¿½ï¿½Æï¿½ï¿½ ï¿½`-*ï¿½:Rï¿½&ï¿½ï¿½'iEï¿½ï¿½>cï¿½ï¿½ï¿½Ë×_OhOï¿½M'Mï¿½O(r)Mï¿½Nï¿½L}ï¿½ï¿½ ï¿½<Nï¿½ï¿½Hï¿½4ï¿½ï¿½ï¿½ZÏï¿½7ï¿½syenÎï¿½.yï¿½ï¿½Dï¿½Iï¿½+ï¿½_-ï¿½]ï¿½xï¿½ï¿½Xï¿½ï¿½ï¿½Âï¿½ï¿½b\$-(ï¿½.ï¿½.ï¿½*ï¿½ï¿½ Vï¿½ï¿½tï¿½ï¿½<e}ï¿½ï¿½Lï¿½Ùªê/ï¿½Xï¿½ï¿½ï¿½[ï¿½;4ï¿½7ï¿½3/4ï¿½ï¿½tï¿½y1/2ï¿½ï¿½Rï¿½ï¿½7Wï¿½(r)%ï¿½ï¿½ï¿½;;Ëº&oï¿½ ï¿½3/4ï¿½ï¿½ï¿½ï¿½w"ï¿½k ï¿½ï¿½ï¿½ ï¿½1/2ï¿½ï¿½Oï¿½\$fOï¿½ï¿½<-ï¿½{ï¿½2~pï¿½ï¿½yï¿½ï¿½ï¿½ï¿½(tm)S6ï¿½ï¿½ï¿½:_1/40ï¿½ï¿½|ï¿½ï¿½ï¿½-vï¿½ï¿½*ï¿½Lp)ï¿½Qï¿½ï¿½ï¿½#ï¿½ï¿½6v\$ï¿½4ï¿½Q ï¿½Qï¿½ï¿½gï¿½ï¿½ï¿½Þ±Ô\$z`'ï¿½a...BG ï¿½ï¿½ï¿½ï¿½Zï¿½71/4(r)ï¿½ï¿½ï¿½(N"*u Õzï¿½&ï¿½Uï¿½ï¿½t'zÂ1ï¿½ï¿½#Uï¿½-kï¿½Mï¿½^ï¿½ï¿½ï¿½qÞ¸Sï¿½ï¿½xA|0ï¿½~"`M8Cï¿½NcGï¿½L\$#ï¿½ï¿½ï¿½ï¿½gIDRiï¿½Ξ(r)-^'3/4-ï¿½ï¿½ï¿½#ï¿½q?&g3aï¿½ï¿½ï¿½ï¿½YDX(r)ï¿½ï¿½Nï¿½Eï¿½ï¿½ï¿½k9ï¿½8ï¿½s-ï¿½'ï¿½zÊï¿½#ï¿½ï¿½'ï¿½ï¿½Ïo"ï¿½K EPW#ï¿½Hï¿½ï¿½Hï¿½ï¿½ï¿½Yï¿½xï¿½Dï¿½dï¿½Tï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½1/4(r)ï¿½ï¿½bï¿½ï¿½>\$ï¿½*ï¿½ï¿½(c)}ï¿½ï¿½ï¿½4ï¿½Jï¿½nï¿½yï¿½Ç¡ï¿½jPnï¿½ï¿½Xï¿½\$ï¿½ï¿½9ï¿½Eï¿½ï¿½ "Rï¿½}ï¿½mï¿½ï¿½]ï¿½ï¿½Bï¿½(c)

2. ## Math Studies I.A

2209000000 6093.3636 Uzbekistan 2,600 64.98 168948 6760000 4222.4004 Venezuela 13,500 73.7 994950 182250000 5431.69 Virgin Islands 14,500 76.86 1114470 210250000 5907.4596 West Bank 2,900 73.46 213034 8410000 5396.3716 Yemen 2,400 62.5 150000 5760000 3906.25 Zimbabwe 200 39.5 7900 40000 1560.25 Total= 1707100 7914.39 130013258 51470030000 559261.8111 r = ,,??-????.-,,??-?? .??-??

1. ## Math IA - Logan's Logo

- represented diagrammatically by the circled red dot. We can now compare my graph to see what the horizontal shift should be. The starting point for the original sine curve is (0, 0) while the starting point for my curve is (-3.0, 0.35). Thus, I determined the value of c=3.0.

2. ## Ib math HL portfolio parabola investigation

(x2 r2x r3x + r2r3) = a(x-r1) (x2 (r2+r3)x + r2r3) = a(x3 - (r2+r3)x2 + r2r3x r1 x2 + r1(r2+r3)x r1r2r3) = a(x3 - (r1+ r2+r3)x2 + (r1+r2 + r1r3 + r2r3)x - r1r2r3) = ax3 - a(r1+ r2+r3)x2 + a(r1+r2 + r1r3 + r2r3)x - a(r1r2r3)

1. ## Shady Areas. In this investigation you will attempt to find a rule to approximate ...

Based on the equation, the percent of variation changes; however, the data is precise at least up to the tenths place. Use other functions to explore the scope and limitations of your general statement. Does it always work? Discuss how the shape of a graph influences your approximation.

2. ## MATH IA- Filling up the petrol tank ARWA and BAO

( which we can see is an inverse function if we replace d with x and the other variables p1 and p2 with constants) Note: The costs to Arwa and Bao per day and the money saved by Bao per day are rounded to two decimal places in the tables MS Excel was used to make the tables.

1. ## MATH IB SL INT ASS1 - Pascal's Triangle

Therefore we can consider the general formula for the denominator as valid. To get the explicit formula of En(r) we only have to put the explicit formulas for the numerator and the denominator together: En(r) = = Xn = 0.5n² + 0.5n Yn(r)

2. ## MATH Lacsap's Fractions IA

The numerators calculated using the statement can be reinforced using Figure 1, Pascal?s triangle. As for the denominator, because of the symmetry observed and the pattern of the difference between the numerator and denominator is still existent. This proves that the general statement is valid and will continue to work. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 