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    Differentiation   
Profit, Cost, and Revenue    Linear Programming      Integration

I will try to add on more each week, so keep looking at this space.
If you find a mistake, please tell me. I may give you a surprise.
 

Integration

We think of integration as the reverse of differentiation and sometimes the integral is referred to as “the antiderivative”.  Now doesn’t that sound impressive.  One thing we must remember is to add a constant every time we integrate without limits. We will get to limits later so don’t panic. When we integrate and get an answer, if we want we can check that we are correct by differentiating the answer and we should get back to where we started.

There are a few basic ways to integrate and I will once again try to explain them as simply as possible.

There are three parts to an integration question
 

                ∫                                           f(x)                                     dx

This sign tells            This is the function       This is the letter that follows
us to integrate          we integrate                 the rules i.e. integrate with respect to

The only part that follows the rules is f(x), the rest just disappears when we integrate.  So lets get on to the different ways we can integrate. Here are the most common ones:

First Way            Simple Integration
Second Way        Integration by substitution
Third Way           Integration by parts
Fourth Way        Integration that ends in a logarithm answer   

First Way       Simple Integration

This is the basic method and it follows this rule:

• Add one to the power
• divide by the new power
   

So lets go through the examples:

Example One
 

     ∫ x ²dx =  ∫ x ²dx      

       The integral is        x    (2+1)      =     x³
                                       (2 +1)                  3

The answer is  x ³ +  c
                          ³

Example Two

     ∫1/ x ²dx     =  ∫  x  ^{- 2} dx

The integral is       x    (^{-  2+ 1} )  =     x ^{-  1}
                              ( - 2+ 1)                  The answer is  -  1   +     c
                           ^x

Be careful when we have a negative index especially when adding.

1/ x is a special case.

Just look what happens when we follow the rule for simple integration:

∫ x  ^{- 1} dx  =       x     ^{-  1+ 1}
                         0

Oh no, we have zero on the bottom of a fraction!!! This should make us think we have to do something else.  Go to the fourth way

Example Three

   ∫5 x ³dx =  5 ∫ x ³dx       We can take numbers outside the integral sign and  multiply the integral by this value.

   The integral is    5     x    ⁽³⁺¹⁾ =   5  x ⁴
                                3 +1                       4

The answer is  5 x⁴ +  c
                                  4

You can only take constants outside the integral sign not variables.

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Second Way    Integration by substitution

This is a strange method as we will use differentiation to help us first.  When we look at the function, it will look more complicated than normal. What we want to do is to make the function look simpler so we can use other methods of integration.

Here is the pattern to follow:

Let the bit that makes it look complicated be replaced by a single letter, usually u.

Write the integral with u and the other x’s.

Differentiate u with respect to x.

Replace all the x’s with u’s even dx with du.

Integrate with respect to u.

Change u’s back to x’s.

Easy as pie!!

So let’s go through the example:

Example Four

  ∫ x ² Ö   (x ³ +1) dx               Please read Ö   (x ³ +1) as the square root of all the bracket
 

Sometimes it is best to choose the part with the highest power of x and you will see why when we differentiate the new letter. We would like the differentiation to be as easy as possible.

Let u = x ³ + 1 this is an easy differentiation so choose this rather than the square root.  There is another reason and you will see this when we differentiate.

So the integral looks like this:

∫ Ö   u     x ²dx                 I call the red part “my leftover x’s”

So now differentiating u,

u = x ³+ 1

du = 3x ^{2            
dx}

There are a few different approaches at this point, choose the one you like the most.

In this bit, just think that you multiply both sides by dx.  It is not mathematically correct but it will do as it works.

du = 3x ²dx

du = 3x ²dx

We need the red part to look exactly like the leftover x’s in the integral
  we have  3x ²dx   after differentiation but we only need   x ²dx`

∫ Ö   u     x ²dx

So   du = 3x ²dx           du =   x ²dx
                                       3

Choosing the term with x ³ gives us a term in x ²when we differentiate and hopefully we can use it to remove the leftover x’s.

Remember we still need the three parts of the integral

∫     f(u)  du     I have written it in terms of u to make it easy to see.

All letters must be the same so either all x’s or all u’s.

∫ Ö   u     du
               3

Now  we can integrate using simple integration, I have taken the number outside the integral sign

  1    ∫   u ^1/2     du
3

 1        u        (^{1/2+ 1}) 
3    (1/2+ 1)

 1        u  ^3/2  
3      3/2

1   2     u ^3/2    
3    3

 2      u  ^3/2    
 9

Now we have to but the x’s back as we started with x’s and we must finish with x’s.

 2   (x³ +1) ^3/2    
 9

The answer is 2    (x³ +1)  ^3/2 + c    
                        9 

Example Five

 _{∫ x e} 3x ² _dx          

Here we are going to let 3x ²be u.

So u = 3 x ²

du = 6x               
dx

du = 6x dx

Now we need the red part to look just like the leftover x’s,

 ∫ e ^u   x dx             du = x dx
                               6

                         ∫ e ^u   du
                                            6
Now  we take the number outside the integral sign,
                                  1      ∫ e ^u   du
                                    6

Remembering back to differentiating exponentials, the function e ^x differentiates to e^x.
So since integration can be considered the opposite of differentiation then e ^x  integrates to e ^x.

So integrating           1      ∫ e ^u   du
                                       6

So the answer is    1    e ^u       it looks like nothing has been done since e^{u     We need to change back to x’s from u’s}

                       1    _e 3x ^{2      
                      6}                
                

The answer is         1    _e 3x ²     +  c    
                               6               

Example Six

           ∫    ( x+3)             dx
               ( x ² +6x) ²

Quite often with this method we will choose the higher power of x, in this case ( x² + 6x) as it will give us something similar to the top term.

Rewriting the integral as ∫  ( x+3)   ( x ² +6x) ^-2   dx

So let

  u = x ² + 6x

du = 2x + 6 = 2(x +3)
dx

dx =     du    
           2(x +3)

So now we will replace dx with      du     
                                                       2(x +3)

∫ ( x+3 ) (u) ^{-  2}   du           =       ∫ ( u) ^{-  2}     du   
                         2(x +3)                                2

So now we can integrate

 1    ∫   ( u) ^{-  2}   du      =       1      (u) ^{-  1}   =  -   1  
2                                        2     (-1)               2u

Now putting the x’s back in

-       1       
 2(x ² + 6x)

 

The answer is    -       1             + c
                             2 ( x ² + 6x)  

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Third Way         Integration by parts

This is often the final resort when all other methods don’t work.  It is not really difficult but you have to choose to differentiate one term whilst integrating the other. Once again without any theory, here is the equation

 ∫  u dv   =   u v -  ∫  v du

Notice that we have a product, u dv, so sometimes you may have to bring the denominator up and make the power negative. Once again, we will be using differentiation to help us as we need du, and integration to find v.  So we will find all the terms individually and then put them into the equation. It is similar to the process to what we did when we have the product or quotient rules in differentiation.

So let’s go through the example:

Example Seven

 ∫  ln x   dx
     x ²

We have to write this as a product so bring x² up to the top.

 ∫  ( ln x)  x ^- 2 dx

Now here is where you have the choice, which one do you integrate? Which one do you differentiate? So look at the two terms carefully, when you differentiate the power gets one less than before so usually it is the x- term. The logarithm term has to be differentiated since we don’t know how to integrate it. The aim is to make the second integral easier than the first integral if not, you may have made the wrong choice so look back to the beginning. The exponential term is equally good at being differentiated or integrated so have a good look at the other term.

Back to our example, now you don’t have to hear me sing but when you see a ln f(x), think  of the song ‘ It had to be you ’. I’m sorry I am a dag!!
So we will have to differentiate ln x and integrate x ^{- 2}

u = ln x                              dv = x ^{-  2}
 

du =    1                               ∫  dv = ∫x ^{-  2}  dx
dx       x

du = 1 dx                         v =    x  ^{-  1      =}  -  x ^{-  1}
         x                                         -  1

So now putting these values into the formula:

 ∫  u dv   =   u v -  ∫  v du

 ∫ ( ln x ) x ^{-  2} dx   =   ln x (-  x ^{-  1}) -    ∫ (-    x ^{-  1}) 1  dx
                                                                              x
                                 =-   ln x   +   ∫x  ^{-  2} dx
                                           x            

 =-  ln x    -   1
         x           x

The answer is        -  ln x   -     1      +  c
                                         x

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Fourth Way        Integration that ends in a logarithm answer

 

∫ x  ^-1 dx  =       x     ^{-  1+ 1}     Obviously not the right way to go!
                         0

Remember this case so what do we do. I know you will hate this but we have to go back to differentiation for the answer. 

Differentiating  Ln (bracket)  = Derivative of the bracket
                                                       Original Bracket

So if we integrate a fraction where the top is the derivative of the bottom, the answer is the logarithm of the bottom.
 

∫  Derivative of the bottom   dx      =  ln ( bottom )
         bottom

So for the troublesome one from before         ∫ 1 dx 
                                                                         The answer is  ln x  + c
 

Sometimes  you may have to multiply top and bottom by a number to make the top the derivative of the bottom or factorise the top. To be very honest, you can do all of this using integration by substitution but I feel this way is a little quicker if you can spot it.

So let’s go through the example:

Example Eight

∫   x           dx
   x ²+ 3

If we look really carefully we will see that x²+ 3  differentiates to 2x which is very close to the top of the fraction. So what I am going to do is multiply the top and bottom by two which is just multiplying by one.

∫   x                2    dx
  ( x ²+ 3)   2

This number two can be taken outside of the integral sign. Just remember that this one is on the bottom so it is one over two i.e. half.

  1    ∫   2 x           dx
 2     ( x ²+ 3)  

Now the top is the derivative of the bottom so the integral is the logarithm of the bottom.

The answer is     1    ln (x ² + 3) +  c
                         2

Example Nine

 ∫   21 + 12x ³   dx
   (7x + x ⁴⁾

If we take 3 out as a common factor we will be left with the derivative of the bottom and once again, we can take this number outside the integral sign.

3 ∫ ( 7 +  4x ³)   dx
     (7x + x ⁴⁾

The answer is    3 ln (7x +x⁴)  +  c

 

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