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Create AccountSo 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...
dfx
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.
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|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 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|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 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.
