Archive

Archive for the ‘Integration’ Category

Integration: the area under a curve

November 9, 2010 Leave a comment

Definite Integration

When we work out a definite integral we end up with a value as opposed to a function

A definite integral is in the form I=\int_a^b f(x)dx

To calculate I

We integrate f(x) (find the inverse derivative)

We substitute in b\text{ and }a into this new function

Then substitute the second value from the first

Notation

\left[F(x)\right]_a^b=F(b)-F(a)

Area under a curve

To find the area under a curve we find the definite integral between the two bounds (ends)

Proof

 

Examples

Find the area (A) under the curve y=4 and the lines x=2\text{ and }x=5

A=\int_2^5 4dx

\implies A=\left[4x\right]_2^5=(4\times 5)-(4\times2)=12

This is to be expected as all we had was rectangle that was 4 high and 3 wide


Find the area (A) under the curve f(x)=3+2x and the lines x=0\text{ and }x=3A=\int_0^3 (3+2x)dx

\implies A=\left[3x+x^2\right]_0^3=(3\times 3+3^2)-(3\times 0+0^2)=18

(note: we do not require the constant of integration because it will be cancelled out in the subtraction)


Find the area (A) under the curve y=3x^2+4x+3 and the lines x=1\text{ and }x=4

The Graph

A=\int_1^3 (3x^2+4x+3)dx

\implies A=\left[x^3+2x^2+3x\right]_1^4=(64+32+12)-(1+2+3)=102

Areas below the x axis

Let’s look at an example

A=\int_2^5 (-4)dx=\left[-4x\right]_2^5=(-4\times 5)-(-4\times2)=-12

Here, we have the same rectangle as in the first example, but this time it is below the x axis.  The result is that the area is given as negative.

We would report the area as positive, but understand that it is below the axis.

Example

Find the area (A) between the x axis the lines x=0\text{ and } x=4 and the curve y=11x-x^2-30

A=\int_0^4 (11x-x^2-30)dx=\left[\frac{11x^2}{2}-\frac{x^3}{3}-30x\right]_0^4

\implies A=(\frac{11\times 16}{2}-\frac{64}{3}-30\times 4)-(0)=53\frac{1}{3}

When areas are above and below the axis

Let’s consider the line y=2x-4 and go from x=0\text{ to } x=4

The graph

A= \int_0^4 (2x-4)dx=\left[x^2-4x\right]_0^4=(16-16)-(0-0)=0

If we look at the graph we can see that there is an area, so what has happened.

Well the area below the axis has cancelled out the area above.

To calculate the desired area we must find the area from 0 to 2 and then from 2 to 4.

A_1= \int_0^2 (2x-4)dx=\left[x^2-4x\right]_0^2=(-4)-(0-0)=-4

So A_1 has an area of 4 (area is positive)

A_2= \int_2^4 (2x-4)dx=\left[x^2-4x\right]_2^4=(16-16)-(4-8)=4

So A_2 has an area of 4

\therefore A=A_1+A_2 = 8

Examples

Find the area between the x axis, x=0,x=2 and the curve f(x)=x^2-1

First we find if the function equals zero between x=0\text{ and }x=2

let f(x)=0\implies x^2-1=0\implies x^2=1\implies x=\pm 1

Since it crosses the axis at x=1 we need to integrate from 0 to 1 and then from 1 to 2.  We will call these areas A_1\text{ and } A_2.

A_1=\int_0^1 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_0^1=(\frac{1}{3}-1)-(0-0)=-\frac{2}{3}

Since A_1>0\implies A_1=\frac{2}{3}

A_2=\int_1^2 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_1^2=(\frac{8}{3}-2)-(\frac{1}{3}-1)=\frac{4}{3}

So the area required = A_1+A_2=\frac{2}{3} + \frac{4}{3} =2


Find the area between the x axis, x=0,x=4 and the curve f(x)=x^2+2x-3

The graph

First we find if the function equals zero between x=0\text{ and }x=4

let f(x)=0\implies x^2+2x-3=0\implies (x-1)(x+3)=0\implies x=-1\text{ or }x=3

Since it crosses the axis at x=1 we need to integrate from 0 to 1 and then from 1 to 4.  We will call these areas A_1\text{ and } A_2.

A_1=\int_0^1 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_0^1=(\frac{1}{3}+1-3)-(0+0-0)=-\frac{5}{3}

A_2=\int_1^4 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_1^4=(\frac{64}{3}+16-12)-(\frac{1}{3}+1-3)=27

So the area required =A_1+A_2=\frac{5}{3}+27=28\frac{2}{3}

Categories: Integration

Integration, the inverse of differentiation

November 3, 2010 Leave a comment

Integration can be used to reverse the effect of differentiation.

The rule for differentiation is: if y=ax^n then \frac{dy}{dx}=anx^{n-1}

So we multiply by the power and then take one off the power.

If we reverse this process we get add one to the power and divide by this new power.

So if \frac{dy}{dx}=ax^n then y=\frac{ax^{n+1}}{n+1}

Lets consider what happens when we differentiate a constant i.e. y=c

This is the same as y=cx^0\Rightarrow \frac{dy}{dx}=0\times cx^{-1}=0

So if we differentiate a constant we get zero which means if we integrate zero we get a constant.

So y=ax^n then \frac{dy}{dx}=anx^{n-1}+c

Examples

\frac{dy}{dx}=8x^3+3x^2-5, what is a general solution for y?

To get y we need to integrate

So y=\frac{8x^4}{4}+\frac{3x^3}{3}-5x+c

\Rightarrow y=2x^4+x^3+5x+c

This is a general solution because c could have any value.

Here I have given c values that range for -1 to 3, but there were an infinite other possibilities.  The important thing to note is that if we differentiate any of these curves they all give the same expression.


The gradient function of a curve is \frac{dy}{dx}=6x^2-3, the curve passes through the point (1,3).

 

What is the particular solution.

Integrating gives y=2x^3-3x+c

Since the curve goes through (1,3) we can substitute in these values to get 3=2-3+c\Rightarrow c=4

This gives the particular solution y=2x^3-3x+4

A particular solution is one where the value of c has been found so that now we can only draw one particular curve.

Notation

\int f(x) dx means integrate f(x).  The dx means that the variable we are considering is x.

Example

Find the general solution of y=\int (3x+4) dx

y=\frac{3x^2}{2}+4x+c

Categories: Integration