Reimann Sum - Midpoint Approximation

Jason76

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Given

15f(x)dx\displaystyle \int^{5}_{1} f(x) dx

interval: [1,5]\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

15f(x)dx\displaystyle \int^{5}_{1} f(x) dx

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

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

a=1\displaystyle a = 1

b=5\displaystyle b = 5

We use

Δ(x)=\displaystyle \Delta(x) = width =ban=512=2\displaystyle = \dfrac{b - a}{n} = \dfrac{5 - 1}{2} = 2

Now we know that each interval is 2\displaystyle 2. So now we need to find the midpoint of each interval using the formula: M1=L1+R22\displaystyle M_{1} = \dfrac{L_{1} + R_{2}}{2}

We know what L1\displaystyle L_{1} and R2\displaystyle R_{2} are for each interval, by looking at a graph we will draw.

Now you would have (x1),(x2)\displaystyle (x_{1}), (x_{2}) which represent each midpoint.

Finally, to find the area under the curve:

Δ(x)[f(x1)+f(x2)]=\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

15f(x)dx\displaystyle \int^{5}_{1} f(x) dx

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

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

a=1\displaystyle a = 1

b=5\displaystyle b = 5

We use

Δ(x)=\displaystyle \Delta(x) = width =ban=512=2\displaystyle = \dfrac{b - a}{n} = \dfrac{5 - 1}{2} = 2

Now we know that each interval is 2\displaystyle 2. So now we need to find the midpoint of each interval using the formula: M1=L1+R22\displaystyle M_{1} = \dfrac{L_{1} + R_{2}}{2}

We know what L1\displaystyle L_{1} and R2\displaystyle R_{2} are for each interval, by looking at a graph we will draw.

Now you would have (x1),(x2)\displaystyle (x_{1}), (x_{2}) which represent each midpoint.

Finally, to find the area under the curve:

Δ(x)[f(x1)+f(x2)]=\displaystyle \Delta(x)[f(x_{1}) + f(x_{2})] = area

Look right :confused:

Looks good. So what did you get for your (x1),(x2)\displaystyle (x_{1}), (x_{2})?
 
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