Video: Expressing a Definite Integral as the Limit of Riemann Sums

Express ∫_(2)^(5) (2π‘₯Β² βˆ’ (5/π‘₯)) dπ‘₯ as the limit of Riemann sums.

03:10

Video Transcript

Express the definite integral between two and five of two π‘₯ squared minus five over π‘₯ with respect to π‘₯ as the limit of Riemann sums.

Remember, if 𝑓 is integrable on the closed interval from π‘Ž to 𝑏, then the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of 𝑓 of π‘₯ 𝑖 times β–³π‘₯. Where β–³π‘₯ is equal to 𝑏 minus π‘Ž over 𝑛 and π‘₯ 𝑖 is equal to π‘Ž plus 𝑖 lots of β–³π‘₯.

In this question, we can see that 𝑓 of π‘₯ is equal to two π‘₯ squared minus five over π‘₯, π‘Ž is the lower limit of our integral, and 𝑏 is the upper limit. So, we’ll begin by establishing whether 𝑓 is actually integrable on the closed interval from π‘Ž to 𝑏. Now, we say that if a function is continuous on a given interval, it’s also integrable on that same interval. So, we need to work out whether the function 𝑓 of π‘₯ equals two π‘₯ squared minus five over π‘₯ is continuous on the closed interval from two to five.

Well, two π‘₯ squared is a polynomial. And we know that polynomial functions are continuous over their entire domain. The function negative five over π‘₯ causes us some issues though. We know that when π‘₯ is equal to zero, five over π‘₯ is five divided by zero, which is undefined. So, the function five over π‘₯ is undefined at the point where π‘₯ equals zero. Luckily though, that’s outside the interval we’re interested in. We’re interested in the closed interval from two to five. And so, we see that five over π‘₯ is a rational function, which is continuous wherever it is defined.

We also know that the sum or difference of two continuous functions is itself a continuous function. So, this means that 𝑓 of π‘₯ is indeed continuous on the closed interval from two to five. And therefore, it must be integrable on that same interval. Great, so, we’ve defined π‘Ž to be two and 𝑏 to be equal to five. And we’ve established that 𝑓 is integrable on the closed interval from two to five. Our next job then is to work out β–³π‘₯.

It’s 𝑏 minus π‘Ž over 𝑛. So, that’s five minus two over 𝑛, which is equal to three over 𝑛. Similarly, we can work out π‘₯ 𝑖. It’s π‘Ž plus 𝑖 lots of β–³π‘₯. So, that’s two plus 𝑖 times three over 𝑛, which is simply two plus three 𝑖 over 𝑛. Now, of course, we need to work out 𝑓 of π‘₯ 𝑖. So, that’s 𝑓 of two plus three 𝑖 over 𝑛. We replace each instance of π‘₯ in our function 𝑓 of π‘₯ with two plus three 𝑖 over 𝑛.

And when we do, we find that 𝑓 of π‘₯ 𝑖 is equal to two times two plus three 𝑖 over 𝑛 squared minus five over two plus three 𝑖 over 𝑛. And we’re now ready to express our definite integral as a limit of Riemann sums. It’s the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛, so these two parts don’t change, of 𝑓 of π‘₯ 𝑖 times β–³π‘₯, which is of course three over 𝑛.

By convention, we bring the three over 𝑛 in front of this nasty expression. And since we’re just being asked to write our definite integral as a limit of Riemann sums, we’re done. We don’t need to evaluate this. It’s the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of three over 𝑛 times two times two plus three 𝑖 over 𝑛 squared minus five over two plus three 𝑖 over 𝑛.

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