Reimann Sum - Midpoint Approximation

Jason76

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Given

\(\displaystyle \int^{5}_{1} f(x) dx\)

interval: \(\displaystyle [1,5]\)

Using two sub-intervals of equal length

What is the midpoint Reimann approximation? Hint :confused:
 
Last edited:
This wasn't isn't too hard to figure out

\(\displaystyle \int^{5}_{1} f(x) dx\)

interval: \(\displaystyle [1,5]\)

\(\displaystyle n = 2\) - because it wants two sub-intervals

\(\displaystyle a = 1\)

\(\displaystyle b = 5\)

We use

\(\displaystyle \Delta(x) =\) width \(\displaystyle = \dfrac{b - a}{n} = \dfrac{5 - 1}{2} = 2\)

Now we know that each interval is \(\displaystyle 2\). So now we need to find the midpoint of each interval using the formula: \(\displaystyle M_{1} = \dfrac{L_{1} + R_{2}}{2}\)

We know what \(\displaystyle L_{1}\) and \(\displaystyle R_{2}\) are for each interval, by looking at a graph we will draw.

Now you would have \(\displaystyle (x_{1}), (x_{2})\) which represent each midpoint.

Finally, to find the area under the curve:

\(\displaystyle \Delta(x)[f(x_{1}) + f(x_{2})] = \) area

Look right :confused:
 
Last edited:
This wasn't isn't too hard to figure out

\(\displaystyle \int^{5}_{1} f(x) dx\)

interval: \(\displaystyle [1,5]\)

\(\displaystyle n = 2\) - because it wants two sub-intervals

\(\displaystyle a = 1\)

\(\displaystyle b = 5\)

We use

\(\displaystyle \Delta(x) =\) width \(\displaystyle = \dfrac{b - a}{n} = \dfrac{5 - 1}{2} = 2\)

Now we know that each interval is \(\displaystyle 2\). So now we need to find the midpoint of each interval using the formula: \(\displaystyle M_{1} = \dfrac{L_{1} + R_{2}}{2}\)

We know what \(\displaystyle L_{1}\) and \(\displaystyle R_{2}\) are for each interval, by looking at a graph we will draw.

Now you would have \(\displaystyle (x_{1}), (x_{2})\) which represent each midpoint.

Finally, to find the area under the curve:

\(\displaystyle \Delta(x)[f(x_{1}) + f(x_{2})] = \) area

Look right :confused:

Looks good. So what did you get for your \(\displaystyle (x_{1}), (x_{2})\)?
 
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