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# Area under a curve using further integration

Started by dfx, Jun 26 2005 08:20 PM

14 replies to this topic

### #1

Posted 26 June 2005 - 08:20 PM

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.

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.

### #2

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.

Half ideas,half quality, half a million pound law suit!

### #3

Posted 26 June 2005 - 08:36 PM

Nah don't think you can do that unless you use partial fractions or something?

### #4

Posted 26 June 2005 - 08:37 PM

Yeh thats what I'm on about.

Half ideas,half quality, half a million pound law suit!

### #5

Posted 26 June 2005 - 10:04 PM

This is covered in Advanced Higher Maths, using the rule:

So for the example you gave:

Hope that doesn't look too scary

So for the example you gave:

Hope that doesn't look too scary

### #6

Posted 26 June 2005 - 10:08 PM

That does look crazzzy to me.. then again, i didnt do ADV maths lol!

### #7

Posted 26 June 2005 - 11:20 PM

Is that Mod x+1 or big brackets x+1 ?

### #8

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?

Half ideas,half quality, half a million pound law suit!

### #9

Posted 27 June 2005 - 02:14 PM

Yes, it's the absolute value of (x+1), since ln isn't defined for negative numbers.

### #10

Posted 27 June 2005 - 11:54 PM

erm... and how would you go about solving it?

I don't understand why its an absolute value, is it cause the asymptote (presumably) never crosses the ordinate into the negative zone?

I don't understand why its an absolute value, is it cause the asymptote (presumably) never crosses the ordinate into the negative zone?

### #11

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

### #12

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.

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### #13

Posted 28 June 2005 - 11:04 PM

Hmmm.. oky doke.. thanks muchos .

Em, what about areas under y = a^x, say 2^x ? I hate learning this trapezium rule so I'm lookin for other methods lol. Thanks.

Em, what about areas under y = a^x, say 2^x ? I hate learning this trapezium rule so I'm lookin for other methods lol. Thanks.

### #14

Posted 29 June 2005 - 08:29 AM

Apparantly

I had to look that up in my calculus textbook though; it's not something I've really met yet.

I had to look that up in my calculus textbook though; it's not something I've really met yet.

### #15

Posted 29 June 2005 - 01:52 PM

You're amazing. Thanks

Now for some show off to the teacher

Now for some show off to the teacher

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