Say you have an equation y = 3/(x+1) and you want to find the area under the curve between the x =2 and x = 6.

So y = 3(x+1) ^-1

Now according to what we were taught in further integration, it would end up as

y = [ 3(x+1) ] / 0 ...

So is there any way round this? Other than using other numerical methods such as the trapezium rule to estimate the area.

## Area under a curve using further integration hmmm...

### Discogirl17

Posted 26 June 2005 - 08:31 PM

You have the equation y=3/(x+1)

so y=3/x+3/1

y=3/x +3

y=3x^-1 +3

y=3+3x

no?

Nope I have no idea. Ignore that whole bit.

so y=3/x+3/1

y=3/x +3

y=3x^-1 +3

y=3+3x

no?

Nope I have no idea. Ignore that whole bit.

### Discogirl17

Posted 27 June 2005 - 01:14 PM

Its mod (x+1) not mod (x) +1

Is that what you were asking?

Is that what you were asking?

### Steve

Posted 28 June 2005 - 10:27 AM

Just in case, the absolute value of a number just means that if it is negative, you make it positive, eg |-2| = 2

The reason its ln|

As an example, consider the equation:

Now for the equation to hold,

If you want to find the area under the curve of

So the area is 1.0986 square units

The reason its ln|

*x*+1| is because the function ln only has positive real numbers in its domain. Think about the graph of*y*=ln(*x*), ie ln(*x*) is undefined for*x*0.As an example, consider the equation:

Now for the equation to hold,

*x*+1 can be -2 or 2. So*x*= -3 or*x*= 1 is the solution.If you want to find the area under the curve of

*y*= 1/*x*however, you just have to sub in the limits as usual into ln|*x*|. For example, the area between*x*= 1 and*x*= 3:So the area is 1.0986 square units

### The Wedge Effect

Posted 28 June 2005 - 04:15 PM

QUOTE(Steve @ Jun 28 2005, 11:27 AM)

Just in case, the absolute value of a number just means that if it is negative, you make it positive, eg |-2| = 2

The reason its ln|

As an example, consider the equation:

Now for the equation to hold,

If you want to find the area under the curve of

So the area is 1.0986 square units

The reason its ln|

*x*+1| is because the function ln only has positive real numbers in its domain. Think about the graph of*y*=ln(*x*), ie ln(*x*) is undefined for*x*0.As an example, consider the equation:

Now for the equation to hold,

*x*+1 can be -2 or 2. So*x*= -3 or*x*= 1 is the solution.If you want to find the area under the curve of

*y*= 1/*x*however, you just have to sub in the limits as usual into ln|*x*|. For example, the area between*x*= 1 and*x*= 3:So the area is 1.0986 square units

*Stares blankly, drool dripping, as if I am braindead*

I'll probs be doing that stuff in uni anyway.