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Sums of Riemann | Exercises, second part

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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