Suppose we want to estimate the area under the curve f(x) = x^2 + 3x - 4 over the interval [1, 4]. We can do this be creating rectangles of equal width (noted as w). Suppose we want to have 4 rectangles, each would be of width (4-1)/4 = 3/4. Using the width of 3/4 we can get the left endpoints of each triangle.
x1 = 1, x2= 1.75, x3 = 2.5, x4 = 3.25 and x5 = 4
f(x1) is the height of the first rectangle using the left endpoint.
f(x2) is the height of the second rectangle using the left endpoint and so on.
Now we take f(x1) = 1 + 3 - 4 = 0
f(x2) = 4.3125
f(x3) = 9.75
f(x4) = 16.3125
To get the area we take w*f(x1) + w*f(x2) + w*f(x3) + w*f(x4). Doing so, we get 22.7815.
Now we can estimate the area again using the right endpoint of each triangle. Those values are
f(x2), f(x3), f(x4) and f(x5)
f(x5) = 24
The estimated area is 22.7815 + 24 = 46.7815.
Using integration on the function f(x), we get (1/3)x^3 + (3/2)x^2 - 4x. Substituting 4 for x, we get 29.333333 Substituting 1 for x, we get -2.166666 Subtracting the values we get the area to be 31.5
Saturday, January 31, 2015
Saturday, January 24, 2015
We can use the anti-derivative or integral to find the area under a curve between two points. For example, suppose we want to find the area under the curve defined by the function f(x) = x^3 + 2x^2 - 4x + 6 between x = 1 and x = 4.
First we integrate the function.
(1/4)x^4 + (2/3)x^3 - 2x^2 + 6x..
Now we evaluate the integral at 4 and then at 1 and subtract.
(1/4)(4)^4 + (2/3)(4)^3 - 2(4)^2 + 6(4) = 98.6666666666
(1/4)(1)^4 + (2/3)(1)^3 - 2(1)^2 + 6(1) = 4.91666666666
Subtract the two to get the area of 93.75
First we integrate the function.
(1/4)x^4 + (2/3)x^3 - 2x^2 + 6x..
Now we evaluate the integral at 4 and then at 1 and subtract.
(1/4)(4)^4 + (2/3)(4)^3 - 2(4)^2 + 6(4) = 98.6666666666
(1/4)(1)^4 + (2/3)(1)^3 - 2(1)^2 + 6(1) = 4.91666666666
Subtract the two to get the area of 93.75
Tuesday, January 20, 2015
There is often confusion between a statistic and a parameter. A statistic is a measure of some attribute of a sample from a population. Examples include mean, standard deviation, variance, quartiles, median, test statistics.
A parameter is a numerical characteristic from a population. According to Wikipedia, a parameter is sometimes taken to be unobservable, and in this case the statistician's task is to infer what they can about the parameter based on observations of random variables distributed according to the probability distribution in question, or, more concretely stated, based on a random sample taken from the population of interest. In other situations, parameters may be fixed by the nature of the sampling procedure used or the kind of statistical procedure being carried out
Source of above paragraph is http://en.wikipedia.org/wiki/Statistical_parameter
So the easy way to distinguish between the two is S-S and P-P. Statistic and sample go together as does parameter and population.
A parameter is a numerical characteristic from a population. According to Wikipedia, a parameter is sometimes taken to be unobservable, and in this case the statistician's task is to infer what they can about the parameter based on observations of random variables distributed according to the probability distribution in question, or, more concretely stated, based on a random sample taken from the population of interest. In other situations, parameters may be fixed by the nature of the sampling procedure used or the kind of statistical procedure being carried out
Source of above paragraph is http://en.wikipedia.org/wiki/Statistical_parameter
So the easy way to distinguish between the two is S-S and P-P. Statistic and sample go together as does parameter and population.
Sunday, January 11, 2015
Suppose a conical tank with a height of 20 feet and a radius of 10 is filled with water at the rate of 5 cubic feet per minute. What will the rate of change of the height be when the level of the tank is 8 feet.
Volume of a cone is 1/3(Pi)r^2(h)
We know dv/dt = 5 and we know that r = (1/2)h. Substitute (1/2)h in for r to get
V = (1/3)Pi(1/2 h)^2(h)
V= (1/3)Pi(1/4)(h^3)
V = (1/12)(Pi)h^3
Now take the derivative to get the rate of change of the volume.
V' = (3/12)(Pi)h^2(dh/dt)
V' = (1/4)Pi(h^2)dh/dt
Substitute 5 for V' and 8 for h to get
5 = (1/4)Pi(64)dh/dt
5 = 50.3dh/dt
.1= dh/dt
Volume of a cone is 1/3(Pi)r^2(h)
We know dv/dt = 5 and we know that r = (1/2)h. Substitute (1/2)h in for r to get
V = (1/3)Pi(1/2 h)^2(h)
V= (1/3)Pi(1/4)(h^3)
V = (1/12)(Pi)h^3
Now take the derivative to get the rate of change of the volume.
V' = (3/12)(Pi)h^2(dh/dt)
V' = (1/4)Pi(h^2)dh/dt
Substitute 5 for V' and 8 for h to get
5 = (1/4)Pi(64)dh/dt
5 = 50.3dh/dt
.1= dh/dt
Saturday, January 10, 2015
When evaluating limits at infinity or negative infinity, there are a couple methods you can use. Look at these three examples.
Lim (x --> infinity) (x^3 + 4x^2 - 3)/(4x^3 + 6x)
You can divide each term by x to the highest power in the problem, which is 3. So divide each term by x^3 to get
Lim (x --> infinity) (1 + 4/x - 3/x^3)/(4 + 6/x^2)
as x tends to infinity, the numerator is just 1 because 4/x and 3/x^3 approach zero. Similarly 6/x^2 approaches zero so the denominator is just 4. Therefore the limit is 1/4.
Another way to do this problem is that since the highest power of x is the same in both numerator and denominator, divide the coefficients of those two terms which is just 1/4.
If the highest power in the numerator is less than the highest power in the denominator, the lim as x tends to infinity or negative infinity is 0. If the highest power in the numerator is greater than the highest power in the denominator, the lim as x tends to infinity or negative infinity is undefined
Lim (x --> infinity) (x^3 + 4x^2 - 3)/(4x^3 + 6x)
You can divide each term by x to the highest power in the problem, which is 3. So divide each term by x^3 to get
Lim (x --> infinity) (1 + 4/x - 3/x^3)/(4 + 6/x^2)
as x tends to infinity, the numerator is just 1 because 4/x and 3/x^3 approach zero. Similarly 6/x^2 approaches zero so the denominator is just 4. Therefore the limit is 1/4.
Another way to do this problem is that since the highest power of x is the same in both numerator and denominator, divide the coefficients of those two terms which is just 1/4.
If the highest power in the numerator is less than the highest power in the denominator, the lim as x tends to infinity or negative infinity is 0. If the highest power in the numerator is greater than the highest power in the denominator, the lim as x tends to infinity or negative infinity is undefined
Tuesday, January 6, 2015
The easy way to tell the nature and number of roots of a quadratic equation is by evaluating its discriminate. The discriminate of a quadratic equation in the form ax^2 + bx + c = 0 is b^2 - 4ac.
If b^2 - 4ac = 0 there is 1 real root
If b^2 - 4ac > 0 there are 2 distinct real roots
If b^2 - 4ac < 0 there are 2 imaginary roots.
If b^2 - 4ac = 0 there is 1 real root
If b^2 - 4ac > 0 there are 2 distinct real roots
If b^2 - 4ac < 0 there are 2 imaginary roots.
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