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 in the range [-3,1] using Riemann sum and then check the result using the corresponding definite integral.
The value of the interval that is more to the left is a=-3 and the other is b=1, let’s consider the equations of \Delta x and x_{i}:
\Delta x = \cfrac{1-(-3)}{n} = \cfrac{4}{n}
x_{i} = -3 + i \cfrac{4}{n} = -3 + 4 \cfrac{i}{n}
Now, let’s propose the sum to be solved, we will substitute \Delta x for what we calculate and we will substitute all 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( -3 + 4 \cfrac{i}{n}\right) \left( \cfrac{4}{n} \right)
Perfect, now we will multiply the two parentheses we have in the summation:
\displaystyle \sum_{i=1}^{n}\left( \cfrac{-12}{n} + 16 \cfrac{i}{n^{2}}\right)
By properties of summations, our sum of the summation will be divided into a sum of summations:
\displaystyle \sum_{i=1}^{n}\cfrac{-12}{n} + \sum_{i=1}^{n}16 \cfrac{i}{n^{2}}
Everything that is not i can leave the summation by properties of the summations because everything that is not i is considered a constant:
\displaystyle -\cfrac{12}{n} \sum_{i=1}^{n} 1 + \cfrac{16}{n^{2}}\sum_{i=1}^{n} i
Now we will apply the sum of a constant and the sum of i that you can find in this article and then proceed to simplify:
-\cfrac{12}{n}n + \cfrac{16}{n^{2}} \left[ \cfrac{n(n+1)}{2} \right] = -12 + \cfrac{16}{n} \left[ \cfrac{n+1}{2} \right]
=-12 + 8 + \cfrac{8}{n} = -4 + \cfrac{8}{n}
Finally we apply the concept of limit, that knowing how it works, all the values that have a denominator n will be equal to zero and the values that do not have n are conserved:
\underset{n \to \infty} \lim \; -4 + \cfrac{8}{n} = -4 + 0 = - 4
Final answer: -4
Let’s check, if the integral of the function f(x) with the limits [-3,1] gives the same result, then our evaluation of the Riemann sum is correct:
\int_{-3}^{1}x \ dx = \left[ \cfrac{x^{2}}{2} \right]_{-3}^{1}
We evaluate:
= \cfrac{1}{2} - \cfrac{9}{2} = -4
Excellent, which means that our result of -4 is correct!
Exercise 2 of the Riemann sums
Evaluate f(x)=x^{2} - x in the range [1,2] using Riemann sum and then check the result using the corresponding definite integral.
Let’s calculate: \Delta x and x_{i}
\Delta x = \cfrac{2 - (1)}{n} = \cfrac{1}{n}
x_{i} = 1 + \cfrac{i}{n}
Now, let’s propose the sum to be solved, we will substitute \Delta x for what we calculated previously 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( 1 + \cfrac{i}{n} \right)^{2} - \left(1 + \cfrac{i}{n} \right) \right] \left(\cfrac{1}{n} \right)
Let’s square the parentheses, remember that it is a perfect square binomial, therefore we will have the following:
\displaystyle \sum_{i=1}^{n} \left[ 1 + \cfrac{2i}{n} + \cfrac{i^{2}}{n^{2}} - 1 - \cfrac{i}{n} \right] \left( \cfrac{1}{n} \right)
Now let’s reduce terms of what is inside the brackets:
\displaystyle \sum_{i=1}^{n} \left[ \cfrac{i}{n} + \cfrac{i^{2}}{n^{2}} \right]\left(\cfrac{1}{n} \right)
We will multiply the bracket with the parenthesis:
\displaystyle \sum_{i=1}^{n} \left( \cfrac{i}{n^{2}} + \cfrac{i^{2}}{n^{3}} \right)
Then we will divide our sum of the sum in a sum of summations:
\displaystyle \sum_{i=1}^{n} \cfrac{i}{n^{2}} + \sum_{i=1}^{n} \cfrac{i^{2}}{n^{3}}
And everything that is not i can leave the sum because everything that is not i is considered a constant:
\displaystyle \cfrac{1}{n^{2}} \sum_{i=1}^{n} i + \cfrac{1}{n^{3}}\sum_{i=1}^{n} i^{2}
If we do not know the sum of i nor the sum of i^{2}, we can consult them by clicking here. After consulting them, it will be as follows:
\cfrac{1}{n^{2}} \left[ \cfrac{n(n+1)}{2} \right] + \cfrac{1}{n^{3}} \left[ \cfrac{n(n+1)(2n+1)}{6} \right]
For the moment, it only remains to simplify terms:
= \cfrac{1}{2n} [n +1] + \cfrac{1}{6n^{2}} [2n^{2} + n + n + 1] =
\cfrac{1}{2} + \cfrac{1}{2n} + \cfrac{1}{3} + \cfrac{1}{3n} + \cfrac{1}{6n^{2}}
Finally we will apply the concept of limit, where everything that has denominator n is equal to zero:
\underset{n \to \infty} \lim \; \left[ \cfrac{1}{2} + \cfrac{1}{2n} + \cfrac{1}{3} + \cfrac{1}{3n} + \cfrac{1}{6n^{2}} \right] = \cfrac{5}{6}
Final answer: \cfrac{5}{6}
Now let’s calculate the integral of the function f(x) in the range of [1,2]:
\int_{1}^{2}(x^{2} - 2) dx = \left[ \cfrac{x^{3}}{3} - \cfrac{x^{2}}{2}\right]_{1}^{2}
We evaluate:
= \cfrac{8}{3} - \cfrac{4}{2} - \cfrac{1}{3} + \cfrac{1}{2} = \cfrac{5}{6}
Excellent, which means that our result of \frac{5}{6} is correct!
Important notes:
- If at the moment of arriving at the part of the evaluation of limits you realize that you have some n with positive exponent in some numerator, some calculation out there was not done well, then you have to review all the operations again.
- It does not hurt to remember it (since sometimes it is forgotten), but remember that the sum of a constant is equal to n multiplied by the constant.
- Courage, it’s easy to do these exercises!
Here you can visit the second part.
Thanks for being in this moment with us : )