Calculus The Definite Integral
Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT

This Time We're Sure!
The Definite Integral


 Say we want to find the area between the line of some equation and the X axis.

 Like maybe, between F(X) = X 2 and the X axis from X = 0 to X = 1
 OK, This is not some silly phoney baloney exercise
 to practice moving symbols around,
 this is very valuable calculus stuff so pay attention here.
 So we have something like:


 This is a weird shape.
 We don't have a formula from geometry for a shape like this.
 We could get a good guess at the actual area of the shape
 by dividing it up into a bunch of little rectangles and adding those areas.
 And calculating the area of a rectangle is easy.


 When we do this, 
 there are little spaces above the rectangles
 that should be included in the area but aren't.
 So we can see right away that our answer won't be exactly right.
 And it will be too low.
 We can also see that if we have more thinner rectangles
 less area at the top of the rectangles will get left out,
 the closer we will get to the right answer.


 If we divide the area into 3 rectangles, each one is 1/3 wide.
 If we divide the area into 50 rectangles, each one is 1/50 wide.
 So in general, 
 If we divide the area into "n" rectangles, each one is 1/n wide.
 So we have:


 If we have 50 rectangles in our 1 unit length of X,
 the first one goes from X = 0 to X = 1/50,
 the second one goes from X = 1/50 to X = 2/50
 the third one goes from X = 2/50 to X = 3/50
 and so on ...


 These numbers are values of X, we can plug them into our equation!

 F(x) = X2

 The function value at each X is the height of that rectangle.
 One More Time ...
 The function value at each X is the height of that rectangle.
 We can actually use the X value on the left side of the rectangle
 or the X value on the right side of the rectangle.
 If we use the left side numbers: 0/50, 1/50, 2/50, 3/50, etc.
 We get these rectangles ...


 If we use the right side of the rectangle height values,
 1/50 for the first, 2/50 for the second, and so on.
 We get:


 If we use the first way (left side X values), the area we get is a bit too small.
 If we use the second way (right side X values), the area we get is a bit too big.
 BUT, if we use lots of tiny rectangles, both answers will be pretty close.
 If we want the area between the curve and the X axis from X = 0 to X = 1
 Think of it as the sum of the area of lots of little rectangles.
 Let's say ... 50 of them (you could use a million if you want)
 Each one is 1/50th of a unit wide.
 To calculate the height of each rectangle, 

 use the X values from one side or the other of each rectangle.

 Since our function is F(x) = X 2, we square each of the X values to get the height.

 The area of each rectangle is the width 1/50 times the height X2  
 So using the left side of each rectangle X values, we get:


 Did you notice that we stopped at 49/50
 because we were using the left side X values???
 We can use the S notation to write this,
 we would have:



 Once we pull all of the constants away, we will be evaluating X2 from X=0 to X=49
 The problem is, that the formulas we have from the last page
 only work when we start at X=1.
 We have two choices.
 We can use right side of the rectangle X values so we go from X=1 to X=50
 We can show you a trick to deal with starting from zero
 Guess which way we're going ...
 Here's the trick
 Since we have X/50 and are starting at 0, we can use (X-1)/50 and start at 1


 Moving constants left:


 Separating the numerator and denominator:


 Moving more constants left:


 Multiply out the right side:


 Separating terms:


 Now we can use those summation formulas
 from the last page:


 n = 50, so:


 Get out your calculator ...



 (42,925 - 2,550 + 50)


 Area 0.3234

 If we do this same process using the right side of the rectangle X values:


 Since we used right side X values, we went to 50/50.
 The S notation for this one is:


 Which can be written as:


 Moving all constants left:


 So the answer is:


 As we figured, this answer is a bit bigger.
 The real answer must be somewhere between these two numbers.
 Most people would probably say:
 "Hey just average the two and that's good enough for me!"
 But of course, math people AREN'T most people.
 So they plodded along looking for really weird equations
 that give us even better answers!
 Aren't we lucky!
 Our general form of the F(X) = X 2 area equation is:


 In the equation, "n" is the number of pieces we cut our horizontal distance into.
 And n is the counting sequence from 1 to whatever we choose n to be.
 The greater the number of rectangles we use
 the closer the area we get will be to the real area.
 The greater number of rectangles we use the smaller each rectangle will be.
 As the number of rectangles gets bigger the width of X (call it  DX) gets smaller.
 If DX gets as absolutely small as it could ever get ...



 That's right. As the number of rectangles approaches infinity, 
 the area gets as close as it can possibly get to the actual area.
 We can even say ...


 So lets solve this.
 Move out the constant and divide up the other stuff ...


 Move the constant to the left...


 Use the rules and evaluate the summation ...



 So the real answer is between the two rectangle answers.
 We KNEW it would be.
 Ya know, this worked and all, 
 but as the equations get more complicated this is going to be a bear
 I wish there were some shortcuts to this limits process.
 Is this like deja vu or what?

   copyright 2005 Bruce Kirkpatrick

Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT