Video: Evaluating the Definite Integral of a Linear Function by Using the Limit of Riemann Sums

Evaluate ∫_(1) ^(3) (βˆ’π‘₯ βˆ’ 2) dπ‘₯ using the limit of Riemann sums.

06:51

Video Transcript

Evaluate the integral from one to three of negative π‘₯ minus two with respect to π‘₯ using the limit of Riemann sums.

We’re asked to evaluate the definite integral of a linear function, and we know we could do this by using the power rule for integration. However, in this question, we’re asked to use the limit of Riemann sums. So let’s start by recalling how we can evaluate the limit of a definite integral by using the limit of Riemann sums. We recall if 𝑓 is an integrable function on the closed interval from π‘Ž to 𝑏, then we can evaluate the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ by evaluating the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of 𝑓 of π‘₯ 𝑖 multiplied by Ξ”π‘₯. And this is the limit as 𝑛 approaches ∞ of the right Riemann sum for 𝑓 of π‘₯. Ξ”π‘₯ will be our interval width. 𝑏 minus π‘Ž divided by 𝑛 and π‘₯ 𝑖 will be our sample points, π‘Ž plus 𝑖 times Ξ”π‘₯.

We want to apply this to the integral given to us in the question, the integral from one to three of negative π‘₯ minus two with respect to π‘₯. First, our function 𝑓 of π‘₯ will be our integrand negative π‘₯ minus two. This means that 𝑓 of π‘₯ is a linear function or a polynomial. And we know polynomials are continuous for all real values of π‘₯. And if a function is continuous on an interval, then that means it must be integrable on that interval. So this means our function will be integrable on any interval. In particular, this means that 𝑓 will be integrable on any closed interval from π‘Ž to 𝑏.

Next, we can find the values of π‘Ž and 𝑏. These will be the upper and lower limits of our integral. So we’ll set our value of π‘Ž equal to one and 𝑏 equal to three. We’re now ready to find an expression for Ξ”π‘₯. We know Ξ”π‘₯ is 𝑏 minus π‘Ž divided by 𝑛. We know 𝑏 is three and π‘Ž is one. So we get Ξ”π‘₯ is three minus one divided by 𝑛, which of course simplifies to give us two divided by 𝑛. We can then use this to find an expression for each of our sample points π‘₯ 𝑖. We know that π‘Ž is equal to one and Ξ”π‘₯ is equal to two over 𝑛, so we get π‘₯ 𝑖 is equal to one plus 𝑖 times two over 𝑛. And we’ll write this as one plus two 𝑖 over 𝑛.

We’re now ready to start formulating the limit of our Riemann sum. We need to find an expression for 𝑓 evaluated at π‘₯ 𝑖. And we’ve already found an expression for π‘₯ 𝑖, so we need to substitute this into our definition for 𝑓 of π‘₯. To evaluate 𝑓 at π‘₯ 𝑖, we need to substitute one plus two 𝑖 over 𝑛 into our function 𝑓 of π‘₯. Remember, 𝑓 of π‘₯ is negative π‘₯ minus two. This gives us negative one times one plus two 𝑖 over 𝑛 minus two. And we can simplify this. We’ll distribute the negative one over our parentheses. This gives us negative one minus two 𝑖 over 𝑛 minus two. And of course, negative one minus two is equal to negative three. Now that we’ve shown that our function 𝑓 is integrable on the closed interval of our integral and we found Ξ”π‘₯, π‘₯ 𝑖, and 𝑓 evaluated at π‘₯ 𝑖, we’re ready to construct our limit.

Substituting in 𝑓 of π‘₯ 𝑖 is negative two 𝑖 over 𝑛 minus three and Ξ”π‘₯ is two over 𝑛, we get the following expression. The integral from one to three of negative π‘₯ minus two with respect to π‘₯ is equal to the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of negative two 𝑖 over 𝑛 minus three multiplied by two over 𝑛. Now, instead of evaluating the definite integral, we can instead evaluate this limit. There’s a few different ways we can approach this. Firstly, we could notice that both two and 𝑛 are constants with respect to our sum, and this means we can just take this outside of our sum. Doing this gives us the limit as 𝑛 approaches ∞ of two over 𝑛 times the sum from 𝑖 equals one to 𝑛 of negative two 𝑖 over 𝑛 minus three.

At this point, we might choose to start evaluating this series. However, we might find it easier to represent this as two separate series. So, by splitting the series in two, we get the limit as 𝑛 approaches ∞ of two over 𝑛 times the sum from 𝑖 equals one to 𝑛 of negative two 𝑖 over 𝑛 plus the sum from 𝑖 equals one to 𝑛 of negative three. And now we can start evaluating a little bit. For example, negative three is a constant. It doesn’t change inside of our series, so the sum from 𝑖 equals one to 𝑛 of negative three will just be 𝑛 copies of negative three. Of course, this evaluates to give us negative three 𝑛. So evaluating our second series gave us negative three 𝑛. Let’s now try and evaluate our first series.

Once again, we can see that there’s a factor of negative two divided by 𝑛. This is a constant in our series, so we can just take it outside of our series. In other words, we’ll rewrite this series as negative two over 𝑛 times the sum from 𝑖 equals one to 𝑛 of 𝑖. Now, we can concentrate on this series itself, the sum from 𝑖 equals one to 𝑛 of 𝑖. These are called the triangular numbers. It’s a well-known result which we can just quote. The sum from 𝑖 equals one to 𝑛 of 𝑖 is equal to one-half 𝑛 times 𝑛 plus one. So we can just write this in for our series. This gives us the limit as 𝑛 approaches ∞ of two over 𝑛 times negative two over 𝑛 multiplied by one-half 𝑛 times 𝑛 plus one minus three 𝑛. And this is a very complicated-looking limit. However, we can simplify it to make it easier.

First, we’ll take the constant factor of two outside of our limit. Next, we want to distribute one over 𝑛 over our outermost set of parentheses. In our first term, we get negative two over 𝑛 multiplied by one over 𝑛 is negative two over 𝑛 squared. And in our second term, we have negative three over 𝑛 multiplied by one over 𝑛. We can cancel the 𝑛’s to just give us negative three. This gives us two times the limit as 𝑛 approaches ∞ of negative two over 𝑛 squared times one-half 𝑛 times 𝑛 plus one minus three. And we can then continue to simplify. For example, in the first term inside of our limit, we can cancel a shared factor of two in the numerator and the denominator. In fact, we can also cancel one of the shared factors of 𝑛 in our numerator and denominator. Doing this, we get two times the limit as 𝑛 approaches ∞ of negative one over 𝑛 times 𝑛 plus one minus three.

The last thing we’ll do is distribute negative one over 𝑛 over our parentheses. Doing this gives us two times the limit as 𝑛 approaches ∞ of negative 𝑛 divided by 𝑛 minus one over 𝑛 minus three. And now we’re ready to just directly evaluate this limit. First, our limit is as 𝑛 is approaching ∞. And since 𝑛 is approaching ∞, 𝑛 will eventually be bigger than zero. This means negative 𝑛 divided by 𝑛 is a constant and is equal to negative one. In our second term of negative one over 𝑛, as 𝑛 is increasing, the denominator is growing without bound. However, our numerator remains constant. This means our second term limits to give us zero. And finally, our third term is a constant, so its limit is just itself, negative three. This gives us two times negative one minus three. And of course, two times negative one minus three is negative eight.

And it’s also worth pointing out we can check our answer by using the power rule for integration to evaluate this definite integral. And if we did this, we would also get negative eight. Therefore, we were able to evaluate the definite integral from one to three of negative π‘₯ minus two with respect to π‘₯ by using the limit of Riemann sums. We got that this definite integral evaluates to give us negative eight.

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