# 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

# Anti-Derivative Proof – 27/02/15

$y=\frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \ln { y } =\ln { \left( \frac { { x }^{ \left( n+1 \right) } }{ \left( n+1 \right) } \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } -\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\ln { \left( { x }^{ \left( n+1 \right) } \right) } +{ C }_{ 2 }\\ \\ \therefore \quad { C }_{ 2 }=-\ln { \left( \left( n+1 \right) \right) } \\ \\ \ln { y } =\left( n+1 \right) \cdot \ln { x } +{ C }_{ 2 }\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { \left( n+1 \right) }{ x } \cdot y\\ \\ \frac { dy }{ dx } =\left( n+1 \right) \cdot { x }^{ -1 }\cdot \frac { 1 }{ \left( n+1 \right) } { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1 }\cdot { x }^{ \left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ -1+\left( n+1 \right) }\\ \\ \frac { dy }{ dx } ={ x }^{ n }\\ \\ \therefore \quad \int { { x }^{ n } } dx=\frac { 1 }{ \left( n+1 \right) } \cdot { x }^{ n+1 }+C$

# Integration Problem – Finding the area of quarter of a circle by integrating…

Integrate:

$\int _{ 0 }^{ r }{ \sqrt { { r }^{ 2 }-{ x }^{ 2 } } } dx$

Say that:

$x=r\sin { \theta }$

So…

$x=r\sin { \theta } \\ \\ x=ru\\ \\ \frac { dx }{ du } =r\\ \\ u=\sin { \theta } \\ \\ \frac { du }{ d\theta } =\cos { \theta } \\ \\ \therefore \quad \frac { dx }{ d\theta } =r\cos { \theta } \\ \\ \therefore \quad dx=r\cos { \theta } d\theta$

Now…

$When\quad x=r,\\ \\ r\sin { \theta =r } \\ \\ \sin { \theta } =1\\ \\ \therefore \quad \theta =\frac { \pi }{ 2 } \\ \\ As\quad \left\{ 0\le \theta \le \frac { \pi }{ 2 } \right\} \\ \\ When\quad x=0,\\ \\ r\sin { \theta } =0\\ \\ \sin { \theta } =0\\ \\ \therefore \quad \theta =0$

So you now have to integrate:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }-{ { r }^{ 2 }\sin ^{ 2 }{ \theta } } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\left( 1-\sin ^{ 2 }{ \theta } \right) } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { { r }^{ 2 }\cos ^{ 2 }{ \theta } } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( { r }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } } { \left( \cos ^{ 2 }{ \theta } \right) }^{ \frac { 1 }{ 2 } }\cdot r\cos { \theta } d\theta \\ \\ =\int _{ o }^{ \frac { \pi }{ 2 } }{ r\cos { \theta } } \cdot r\cos { \theta } d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \cos ^{ 2 }{ \theta } d\theta$

But first, realise that:

$\cos { \left( \theta +\theta \right) } =\cos { \theta } \cos { \theta } -\sin { \theta } \sin { \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\sin ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -\left( 1-\cos ^{ 2 }{ \theta } \right) \\ \\ \cos { 2\theta } =\cos ^{ 2 }{ \theta } -1+\cos ^{ 2 }{ \theta } \\ \\ \cos { 2\theta } =2\cos ^{ 2 }{ \theta } -1\\ \\ 2\cos ^{ 2 }{ \theta } =\cos { 2\theta } +1\\ \\ \cos ^{ 2 }{ \theta } =\frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 }$

So you now integrate:

$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { r }^{ 2 } } \left( \frac { 1 }{ 2 } \cos { 2\theta } +\frac { 1 }{ 2 } \right) d\theta \\ \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 1 }{ 2 } } { r }^{ 2 }\cos { 2\theta } +\frac { 1 }{ 2 } { r }^{ 2 }\quad d\theta$

And:

$If\quad p=\frac { 1 }{ 4 } { r }^{ 2 }\sin { 2\theta } =\frac { 1 }{ 4 } { r }^{ 2 }u\\ \\ \frac { dp }{ du } =\frac { 1 }{ 4 } { r }^{ 2 }\\ \\ u=\sin { 2\theta } =\sin { q } \\ \\ \frac { du }{ dq } =\cos { q } =\cos { 2\theta } \\ \\ q=2\theta ,\quad \frac { dq }{ d\theta } =2\\ \\ \frac { du }{ d\theta } =2\cos { 2\theta } \\ \\ \therefore \quad \frac { dp }{ d\theta } =\frac { 1 }{ 4 } { r }^{ 2 }\cdot 2\cos { 2\theta } \\ \\ =\frac { 1 }{ 2 } { r }^{ 2 }\cos { 2\theta }$

Therefore:

Now, to get the area of a circle, you multiply the final result by 4.

$4\cdot \frac { 1 }{ 4 } \pi { r }^{ 2 }\\ \\ =\pi { r }^{ 2 }\\ \\ \therefore \quad A=\pi { r }^{ 2 }$

# Derivative of y=cotx

$y=cotx=\frac { cosx }{ sinx } =\frac { u }{ v } ,\\ \\ If\quad y=\frac { u }{ v } ,\quad \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } } .\\ \\ u=cosx,\quad \therefore \quad \frac { du }{ dx } =-sinx\\ v=sinx,\quad \therefore \quad \frac { dv }{ dx } =cosx,\quad { v }^{ 2 }=\sin ^{ 2 }{ x } \\ \\ So:\\ \\ \frac { dy }{ dx } =\frac { sinx\cdot \left( -sinx \right) -cosxcosx }{ \sin ^{ 2 }{ x } } \\ \\ =\frac { -\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } =\frac { -\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } -\frac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } \\ \\ =-1-\cot ^{ 2 }{ x } =-{ cosec }^{ 2 }x\\ \\ \therefore \quad If\quad y=cotx,\quad \frac { dy }{ dx } =-{ cosec }^{ 2 }x\\ \\ \\$

# Quick Way To Derive The Integration By Parts Formula

Firstly, you need to know what the product rule is:

$If\quad y=u\cdot v,\quad \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } .$

Then…

$\frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ u\frac { dv }{ dx } =\frac { dy }{ dx } -v\frac { du }{ dx }$

Now integrate each term with respect to x:

$\int { u\frac { dv }{ dx } } dx=\int { \frac { dy }{ dx } } dx-\int { v\frac { du }{ dx } } dx$

Leaving:

$\int { u\frac { dv }{ dx } } dx=y-\int { v\frac { du }{ dx } } dx\\ \\ As\quad y=u\cdot v,\\ \\ \int { u\frac { dv }{ dx } } dx=uv-\int { v\frac { du }{ dx } } dx$

# Derivate Of y=cosx Proof

Prove that:

$If\quad f(x)=cosx,\quad f$

$\lim _{ \delta x\rightarrow 0 }{ \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cos\left( x+\delta x \right) -cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosxcos\delta x-sinxsin\delta x-cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) -sinxsin\delta x }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) }{ \delta x } } -\frac { sinxsin\delta x }{ \delta x } \\ \\ =0-sinx=-sinx\\ \\ As:\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { \left( cos\delta x-1 \right) }{ \delta x } } =0\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { sin\delta x }{ \delta x } } =1$

# Tricky Edexcel C4 Examination Question Solved (January 2008) Paper

Question (8):

Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm^3 per second and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm^2.

a) Show that at time seconds, the height h cm of liquid in the cylinder satisfies the differential equation:

$\frac { dh }{ dt } =0.4-k\sqrt { h }$, where is a positive constant.

Solution:

Well firstly let’s draw a little diagram to depict what is actually happening:

As you can see, dV/dt is the rate at which the volume of liquid in the cylinder is changing over time. You have to remember that liquid is being poured into  the large vertical cylinder at a constant rate of 1600 cm^3 per second. What you also have to note is that the vertical cylinder is losing liquid at a rate proportional to the square root of the height already in the cylinder. Therefore we have $\frac { dV }{ dt } =1600-C\sqrt { h }$.

Now the area of the circular cross section should be: $A=\pi { r }^{ 2 }$. However, we are told that A=4000. Knowing that the Volume of the cylinder is $V=\pi { r }^{ 2 }h$, we can transform the volume formula to V=4000h.

Ok, so how can we get dh/dt? This is the formula we can use to derive it:

$\frac { dh }{ dt } =\frac { dh }{ dV } \cdot \frac { dV }{ dt }$

We know that:

$V=4000h,\\ \\ \therefore \quad \frac { dV }{ dh } =4000,\quad and\quad as\quad \frac { dh }{ dV } =\frac { 1 }{ \frac { dV }{ dh } } ,\\ \frac { dh }{ dV } =\frac { 1 }{ 4000 }$.

We also know that:

$\frac { dV }{ dt } =1600-C\sqrt { h }$

So what we get is:

$\frac { dh }{ dt } =\frac { 1 }{ 4000 } \cdot \left( 1600-C\sqrt { h } \right) \\ \\ =\frac { 1600 }{ 4000 } -\frac { C }{ 4000 } \sqrt { h } \\ \\ =0.4-k\sqrt { h } ,\\ \\ \therefore \quad k=\frac { C }{ 4000 }$.

—————————————————————————

b) When h=25, water is leaking out of the hole at 400 cm^3 per second. Show that k=0.02.

Solution:

Ok, with this information we can say that:

$\frac { dV }{ dt } =1600-400$

Moving on, we can determine that:

$\frac { dV }{ dt } =1600-400,\quad but\quad \frac { dV }{ dt } =1600-C\sqrt { h } .\\ \\ Since,\quad h=25,\quad and\quad C\sqrt { h } =400,\\ \\ 5C=400,\quad \therefore \quad C=80.\\ \\ But\quad k=\frac { C }{ 4000 } =\frac { 80 }{ 4000 } ,\\ \\ \therefore \quad k=0.02.\\ \\$

—————————————————————

c) Separate the variables of the differential equation:

$\frac { dh }{ dt } =0.4-0.02\sqrt { h } \\ \\$,

to show that the time taken to fill the cylinder from empty to a height of 100cm is given by:

$\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

Solution:

$\frac { dh }{ dt } =\frac { 2 }{ 5 } -\frac { \sqrt { h } }{ 50 } =\frac { 20 }{ 50 } -\frac { \sqrt { h } }{ 50 } \\ \\ =\frac { 20-\sqrt { h } }{ 50 } \\ \\ \frac { dh }{ dt } =\frac { 20-\sqrt { h } }{ 50 } \\ \\ dh=\frac { 20-\sqrt { h } }{ 50 } dt\\ \\ 50dh=20-\sqrt { h } dt\\ \\ \frac { 50 }{ 20-\sqrt { h } } dh=1dt\\ \\ \int { 1dt=\int { \frac { 50 }{ 20-\sqrt { h } } } } dh\\ \\ \therefore \quad t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

———————————————————————

d) Using the substitution:

$h={ \left( 20-x \right) }^{ 2 }\\ \\$, or otherwise, find the exact value of:

$\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

Solution:

$If\quad h={ \left( 20-x \right) }^{ 2 },\quad \\ \\ { h }^{ \frac { 1 }{ 2 } }=\sqrt { h } ={ \left[ { \left( 20-x \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=\left( 20-x \right) .\\ \\ If\quad h={ \left( 20-x \right) }^{ 2 }={ p }^{ 2 },\quad \frac { dh }{ dp } =2p,\quad p=20-x\quad \therefore \quad \frac { dp }{ dx } =-1\\ \\ So\quad \frac { dh }{ dx } =-2\left( 20-x \right) ,\quad \therefore \quad dh=-2\left( 20-x \right) dx\\ \\ When\quad h=100,\quad { \left( 20-x \right) }^{ 2 }=100,\quad \therefore \quad x=10.\\ When\quad h=0,\quad { \left( 20-x \right) }^{ 2 }=0,\quad \therefore \quad x=20.\$

Therefore, we need to integrate:

$\int _{ 20 }^{ 10 }{ \frac { 50 }{ 20-\left( 20-x \right) } \cdot \frac { -2\left( 20-x \right) }{ 1 } } dx=\int _{ 20 }^{ 10 }{ \frac { -100\left( 20-x \right) }{ x } } dx\\ \\ =\int _{ 20 }^{ 10 }{ \frac { -2000+100x }{ x } } dx=\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +\frac { 100x }{ x } dx\\ \\ =\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx.\\ \\ But\quad as\quad -\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ b }^{ a }{ f\left( x \right) dx } ,\\ \\ \int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx=\int _{ 10 }^{ 20 }{ \frac { 2000 }{ x } } -100dx\\ \\ ={ \left[ 2000\ln { x } -100x \right] }_{ 10 }^{ 20 }\\ \\ =2000\ln { 20 } -2000-\left( 2000\ln { 10-1000 } \right) \\ \\ =2000\ln { 20 } -2000-2000\ln { 10 } +1000\\ \\ =2000\left( \ln { 20-\ln { 10 } } \right) -2000+1000\\ \\ =2000\ln { 2-1000 }$

——————————————————————–

e) Hence find the time taken to fill the cylinder from empty to a height of 100cm, giving your answer in minutes and seconds to the nearest second.

Solution:

Remember that:

$t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh$.

As this is the case,

$t=2000\ln { 2 } -1000=386\quad seconds=6\quad minutes\quad and\quad 26\quad seconds.$

# Coming Up With The Formula For Areas Underneath Curves

$Area\quad PMTN<\delta A

# Proving The Product Rule

$If\quad y=u\cdot v,\quad \left( y+\delta y \right) =\left( u+\delta u \right) \left( v+\delta v \right) \\ \\ y+\delta y=uv+u\delta v+v\delta u+\delta u\delta v\\ \\ but\quad y=uv,\\ \\ \delta y=u\delta v+v\delta u+\delta u\delta v\\ \\ \frac { \delta y }{ \delta x } =u\frac { \delta v }{ \delta x } +v\frac { \delta u }{ \delta x } +\frac { \delta u }{ \delta x } \delta v\\ \\ But\quad as\quad \delta x\rightarrow 0,\quad \frac { \delta y }{ \delta x } \rightarrow \frac { dy }{ dx } ,\quad \frac { \delta v }{ \delta x } \rightarrow \frac { dv }{ dx } ,\quad \frac { \delta u }{ \delta x } \rightarrow \frac { du }{ dx } \quad and\quad \delta v\rightarrow 0.\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx }$