# Derive the formula to find areas underneath curves

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

In this diagram what you will discover is that:

• A length a exists, which starts at the origin O and ends at a;
• A length x exists, which starts at the origin O and ends at x;
• A length x+𝛿x exists, which starts at the origin O and ends at x+𝛿x;
• A length 𝛿x exists, which starts at x and ends at x+𝛿x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+𝛿y exists, which starts at the origin O and ends at y+𝛿y;
• A height 𝛿y exists, which starts at y and ends at y+𝛿y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve called 𝛿A which commences at x and ends at x+𝛿(Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called 𝛿A);
• There is a rectangle that exists called QRUT. It has an area which is y𝛿x;
• There is a rectangle that exists called PRUS. It has an area which is (y+𝛿y)𝛿x;
• 𝛿A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT < 𝛿A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y𝛿x < 𝛿A < (y+𝛿y)𝛿x

Now, check out what happens when we divide all the elements of this expression by 𝛿x:

What we end up with is…

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

This is because as 𝛿x approaches 0, 𝛿y approaches 0 leaving (𝛿A)/(𝛿x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

As a consequence, this ultimately means that:

$y=\frac { dA }{ dx }$

This is incredibly significant, because if we then integrate both sides of this equation, we get:

$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$

And…

$A=\int { ydx } =F\left( x \right) +C$

Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.

Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

$A=\int { ydx } =F\left( x \right) +C$

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the area A underneath the curve is 0, watch what happens… Look at what we get…

$O=F\left( a \right) +C$

Which means that:

$C=-F\left( a \right)$

Hence, we can conclude that:

$A=\int { ydx=F\left( x \right) } -F\left( a \right)$

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b

This is probably the formula you’re most familiar with…

$A=\int _{ a }^{ b }{ ydx=F\left( b \right) } -F\left( a \right) ={ \left[ F\left( x \right) \right] }_{ a }^{ b }$

Which is the formula which can be used to find areas underneath curves.

If you are still confused and would like to go through this proof once again, please watch my video below…

Related:

Trapezium Rule Formula – Derivation

# Deriving the Taylor Series from scratch

[Please note: In order to derive the Taylor Series, you will need to understand how to differentiate. If you know how to differentiate, finding the Taylor Series won’t be much of a problem. You also need to know that 0!=1, 1!=1, 2!=2, 3!=6, x^0=1, x^1=x.]

In this post I will be demonstrating how one can produce the Taylor Series from absolute scratch.

First of all, let’s look at the diagram above. Now, let’s suppose that the equation of the function above is:

$f\left( x+a \right) ={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...$

Ok, so we have the equation for the function, however, it isn’t complete. C_0, C_1, C_2, C_3 etc are hidden constants. This means that our second task will be to discover these constants. We need to discover these constants to find the complete equation of the function so that we can arrive at the Taylor Series. Fortunately, this task won’t be too difficult. Let me show you how C_0, C_1, C_2, C_3 etc can be found fairly easily…

When x=0:

${ f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }\quad \therefore \quad \frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( a \right) ={ C }_{ 0 }$

Now:

${ f }^{ \left( 1 \right) }\left( x+a \right) ={ C }_{ 1 }+2{ C }_{ 2 }x+3{ C }_{ 3 }{ x }^{ 2 }+...$

When x=0:

${ f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }\quad \therefore \quad \frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( a \right) ={ C }_{ 1 }$

Also:

${ f }^{ \left( 2 \right) }\left( x+a \right) =2{ C }_{ 2 }+6{ C }_{ 3 }x+...$

When x=0:

${ f }^{ \left( 2 \right) }\left( a \right) =2{ C }_{ 2 }\quad \therefore \quad \frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( a \right) ={ C }_{ 2 }$

And, finally:

${ f }^{ \left( 3 \right) }\left( x+a \right) =6{ C }_{ 3 }+...$

When x=0:

${ f }^{ \left( 3 \right) }\left( a \right) =6{ C }_{ 3 }\quad \therefore \quad \frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( a \right) ={ C }_{ 3 }$

Alright, so now that we have discovered the hidden constants C_0, C_1, C_2 and C_3, our third task is to write down the complete equation of the function f(x+a). Thanks to the information we have above, the fact that x^0=1 and x^1=x, plus our ability to spot patterns, we will be able to do this quite quickly…

[*Image can be seen here if it appears to be too small on this page.]

And it turns out that the equation we have just above is the Taylor Series function. It can be simplified to look like this…

What is also interesting is that if we transform a=0, we get the Maclaurin Series function which can be used to discover formulas for things such as e^x.

If you have any questions regarding this post, please leave your comments below. Once again, thanks for stopping by! 🙂

Related:

How to derive Euler’s Identity using the Maclaurin Series

# Function Rules

$fg\left( x \right) =f\left( g\left( x \right) \right) \\ \\ ff\left( x \right) =f\left( f\left( x \right) \right) ={ f }^{ 2 }\left( x \right) \\ \\ gf\left( x \right) =g\left( f\left( x \right) \right) \\ \\ gg\left( x \right) =g\left( g\left( x \right) \right) ={ g }^{ 2 }\left( x \right)$

# Even And Odd Functions

An even function exists when $f\left( x \right) =f\left( -x \right)$ for all values of x.

The graph of an even function must be symmetrical about the y-axis.

Examples of even functions have been listed below:

$f\left( x \right) ={ x }^{ 2 }$

$f\left( x \right) ={ x }^{ 4 }$

$f\left( x \right) =cosx$

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An odd function exists when $f\left( x \right) =-f\left( -x \right)$ for all values of x.

Graphs of odd functions should have 180 degrees rotational symmetry about the origin (0,0).

Examples of odd functions are listed below:

$f\left( x \right) =x$

$f\left( x \right) ={ x }^{ 3 }$

$f\left( x \right) =sinx$

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Most functions are neither even or odd.

An example of a function that is neither even or odd would be:

$f\left( x \right) ={ x }^{ 2 }+\frac { 1 }{ x }$