For the resolution of the exercises presented below, we have to apply the following Riemann sum:
\displaystyle \sum_{i=1}^{n} f(x_{i})\Delta x
Where \Delta x and x_{i} are calculated as follows:
\Delta x = \cfrac{b - a}{n}
x_{i} = a + i\Delta x
Maybe we will need some formulas of summations that you can find by clicking here.
Exercise 1 of the Riemann sums
Evaluate f(x)=(x^{3}-1) in the range [0,1] using Riemann sum and then check the result using the corresponding definite integral.
The value of the interval that is furthest to the left is a=0 and the other is b=1, let’s consider the equations of \Delta x and x_{i}:
\Delta x = \cfrac{1 - (0)}{n} = \cfrac{1}{n}
x_{i} = a + i + \Delta x = 0 + \cfrac{i}{n} = \cfrac{i}{n}
Now, let’s consider the sum to be solved, we will substitute \Delta x for what we calculate and we will substitute all the x of the function f(x) for what we have calculated in x_{i} previously:
\displaystyle \sum_{i=1}^{n}f(x_{i})\Delta x = \sum_{i=1}^{n}\left[ \left(\cfrac{i}{n} \right)^{3} - 1 \right] \left( \cfrac{1}{n} \right)=
Let’s raise the cube and then multiply the parenthesis with the bracket:
\displaystyle \sum_{i=1}^{n} \left[\cfrac{i^{3}}{n^{3}} - 1 \right] \left(\cfrac{1}{n} \right) = \sum_{i=1}^{n} \left( \cfrac{i^{3}}{n^{4}} - \cfrac{1}{n} \right)
Now let’s divide our subtraction of the summation into a subtraction of summations:
\displaystyle \sum_{i=1}^{n}\cfrac{i^{3}}{n^{4}} - \sum_{i=1}^{n}\cfrac{1}{n}
We will remove everything that is not i from the summation for properties of the summation because everything that is not i is considered a constant:
\displaystyle \cfrac{1}{n^{4}} \sum_{i=1}^{n} i^{3} - \cfrac{1}{n} \sum_{i=1}^{n} 1
By applying summation formulas, we will obtain the following:
\cfrac{1}{n^{4}} \left[ \cfrac{n^{2}(n+1)^{2}}{4} \right] - \cfrac{1}{n} [n]
Now we’re just going to simplify as much as we can:
= \cfrac{1}{4n^{2}}\left[ (n+1)^{2} \right] - 1 = \cfrac{1}{4n^{2}}(n^{2} + 2n + 1) - 1
= \cfrac{1}{4} + \cfrac{1}{2n} + \cfrac{1}{4n^{2}} - 1
We only need to apply the concept of limit to obtain our final result:
\underset{n \to \infty} \lim \; \left[ \cfrac{1}{4} + \cfrac{1}{2n} + \cfrac{1}{4n^{2}} - 1 \right] = -\cfrac{3}{4}
Final answer: -\cfrac{3}{4}
Let’s calculate the integral of the function f(x) in the range of [0,1]:
\displaystyle \int_{0}^{1}(x^{3} - 1) dx = \left[ \cfrac{x^{4}}{4} - x \right]_{0}^{1}
We evaluate:
=\cfrac{1}{4} - 1 = -\cfrac{3}{4}
Excellent, which means that our result of -\frac{3}{4} is correct!
Exercise 2 of the Riemann sums
Evaluate f(x)=4 in the range [3,6] using Riemann sum and then check the result using the corresponding definite integral. This Riemann summation exercise is very simple.
Let’s calculate \Delta x and x_{i}:
\Delta x = \cfrac{6 - (3)}{n} = \cfrac{3}{n}
x_{i} = 3+ i \Delta x = 3 + \cfrac{3i}{n}
Now, let’s consider our summation, replace the value of \Delta x, and because we do not have x in the function of f (x) we will keep the value of 4, but let’s stop talking and consider our equations so that it is better understood:
\displaystyle \sum_{i=1}^{n}f(x_{i})\Delta x = \sum_{i=1}^{n} f\left( 3 + \cfrac{3i}{n}\right) \left(\cfrac{3}{n} \right) = \sum_{i=1}^{n} [4]\left(\cfrac{3}{n} \right)
We multiply the bracket with the parenthesis:
\displaystyle \sum_{i=1}^{n} \cfrac{12}{n}
And we get everything that is not i from the sum because everything that is not i is considered a constant:
\displaystyle \cfrac{12}{n} \sum_{i=1}^{n} 1 = \cfrac{12}{n} [n] = 12
Now we apply the concept of limit:
\underset{n \to \infty} \lim \; 12 = 12
Final answer: 12
Finally, calculate the integral of the function f(x) in the range of [3,6]:
\displaystyle \int_{3}^{6}4 \ dx = \left[ 4 x \right]_{3}^{6}
We evaluate:
4(6) - 4(3) = 24 - 12 = 12
Excellent, which means that our result of 12 is correct!
Here you can visit part 1.
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