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Example Questions
Example Question #181 : Midpoint Riemann Sums
Utilize the method of midpoint Riemann sums to approximate using three midpoints.
A Riemann sum integral approximation over an interval with subintervals follows the form:
It is essentially a sum of rectangles each with a base of length and variable heights , which depend on the function value at a given point .
We're asked to approximate
So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are
Now, the complex number in the exponent may've been a point of curiosity before a calculator was used to find the result. However, this is a good opportunity to point out a unique mathematical identity: Euler's Identity. Euler's Identity, or rather a facet of it states:
or
In the case of our problem sum, two of our three elements represent a function of this form:
Mathematical beauty.
Example Question #181 : How To Find Midpoint Riemann Sums
Utilize the method of midpoint Riemann sums to approximate using three midpoints.
A Riemann sum integral approximation over an interval with subintervals follows the form:
It is essentially a sum of rectangles each with a base of length and variable heights , which depend on the function value at a given point .
We're asked to approximate
So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are
Example Question #181 : How To Find Midpoint Riemann Sums
Utilize the method of midpoint Riemann sums to approximate using three midpoints.
A Riemann sum integral approximation over an interval with subintervals follows the form:
It is essentially a sum of rectangles each with a base of length and variable heights , which depend on the function value at a given point .
We're asked to approximate
So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are
Example Question #182 : Midpoint Riemann Sums
Utilize the method of midpoint Riemann sums to approximate using three midpoints.
A Riemann sum integral approximation over an interval with subintervals follows the form:
It is essentially a sum of rectangles each with a base of length and variable heights , which depend on the function value at a given point .
We're asked to approximate
So the interval is , the subintervals have length , and since we are using the midpoints of each interval, the x-values are
Example Question #1211 : Calculus
Calcuate the Midpoint Riemann sum of the function
on the interval divided into three sub-intervals.
The interval divided into three sub-intervals gives rectangles with the vertices of the bases at
For the Midpoint Riemann sum, we need to find the rectangle heights which come from the midpoint of the sub-intervals, or , , and .
Because each sub-interval has a width of , the Midpoint Riemann sum is
Example Question #181 : Functions
Calcuate the Midpoint Riemann sum of the function
on the interval divided into four sub-intervals.
The interval divided into four sub-intervals gives rectangles with the vertices of the bases at
For the Midpoint Riemann sum, we need to find the rectangle heights which come from the midpoint of the sub-intervals, or , , , and .
Because each sub-interval has a width of , the Midpoint Riemann sum is
Example Question #187 : Differential Functions
Calculate the area between , , , and the x-axis using Riemann Midpoint sums.
Riemann Sums are a way to approximate the area under a curve. While taking an integral gives you an exact area, using Riemann Sums is a way to get a fast estimation given a function that might be more difficult to integrate. There are many different ways to use this technique. Given some number, n (which as it increases gives you better approximations), we can split the area under the curve into n rectangles. The height of each rectangle depends on what version of Riemann Sums you are using. Three such versions are left endpoint, right endpoint, and midpoint. The formula itself is simple, say we have a curve bounded in [a,b]. Then we can approximate the area using the following equation
.
Where the are a list of midpoints for each rectangle and which is the width of each rectangle.
First we calculate the length of each subinterval
.
Now we look at the first rectangle.
It is in between 5 and 7 so the midpoint of the two is 6.
Now we easily know the midpoint of all 5 of the rectangles
.
Thus the area is thus
which is simply
.