mellamokb
Joined: 10 May 2007
Posts: 124
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 Posted: October 22, 2007 9:27 PM Post subject: |
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Hi bigger,
I'm assuming your expression is in y1(x), if not, adjust my code accordingly.
For the minimum and maximum values, you can have the calculator determine the exact values using the following functions:
Note: OUT denotes the calculators output, and <= is the less than/equal to key (single character).
| Code: |
fMin(y1(x),x)|0<=x and x<=4
OUT: x=0 or x=4
fMax(y1(x),x)|0<=x and x<=4
OUT: x=9/4
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For the mean, I believe it is just the area divided by the width of the interval, which can be done as follows:
Note: _/ is the integral [F3]+[2] or [2ND]+[7].
| Code: |
_/(y1(x),x,0,4)/(4-0)
OUT: 1/4
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The median involves finding the point at which half of the area is to the left and half of the area is to the right. This is equivalent to solving two equal integrals, the first having bound [0, b] and the second having bound [b, 4]:
| Code: |
solve(_/(y1(x),x,0,b)=_/(y1(x),x,b,4),b)|0<=b and b<=4
OUT: b=2.1641353083
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However, since we know the area to be _/(y1(x),x,0,4)=1, then it is equivalent to finding the point b where the area is 1/2, because that would be half the area:
| Code: |
solve(_/(y1(x),x,0,b)=1/2,b)|0<=b and b<=4
OUT: b=2.1641353083
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Hope this helps!
~ mellamokb |
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