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midpoint rule

midpoint rule

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midpoint rule

Postby Integr88 » Thu 31 March, 2011 21:34

using midpoint rule solve x^2+1 on the interval [0,2]

the formula for the midpoint is `M=((x_1+x_2)/2,(y_1+y_2)/2)`

have two point `x_1=0` and `x_2=2`. so how do we get the `y_1` and `y_2`?

simply sub in the values into the equation which is `y=x^2+1`

first to find `y_1` we use `x_1` which is 0
`y_1= 0^2+1=1`
`y_1=1`

first to find `y_2` we use `x_2` which is 2
`y_2=2^2+1=5`
`y_2=5`

Now we use use the `x_1, x_2, y_1, y_2` into our midpoint equation `M=((x_1+x_2)/2,(y_1+y_2)/2)`
`M=((0+2)/2,(1+5)/2)=1/2,6/2`

`:.` so we found the midpoint to be `(1/2, 6/2)
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