Video Transcript
Express the definite integral between two and five of two 𝑥 squared minus five over 𝑥 with respect to 𝑥 as the limit of Riemann sums.
Remember, if 𝑓 is integrable on the closed interval from 𝑎 to 𝑏, then the definite integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 with respect to 𝑥 is equal to the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of 𝑓 of 𝑥 𝑖 times △𝑥. Where △𝑥 is equal to 𝑏 minus 𝑎 over 𝑛 and 𝑥 𝑖 is equal to 𝑎 plus 𝑖 lots of △𝑥.
In this question, we can see that 𝑓 of 𝑥 is equal to two 𝑥 squared minus five over 𝑥, 𝑎 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 𝑎 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 𝑥 equals two 𝑥 squared minus five over 𝑥 is continuous on the closed interval from two to five.
Well, two 𝑥 squared is a polynomial. And we know that polynomial functions are continuous over their entire domain. The function negative five over 𝑥 causes us some issues though. We know that when 𝑥 is equal to zero, five over 𝑥 is five divided by zero, which is undefined. So, the function five over 𝑥 is undefined at the point where 𝑥 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 𝑥 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 𝑥 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 𝑎 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 △𝑥.
It’s 𝑏 minus 𝑎 over 𝑛. So, that’s five minus two over 𝑛, which is equal to three over 𝑛. Similarly, we can work out 𝑥 𝑖. It’s 𝑎 plus 𝑖 lots of △𝑥. 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 𝑥 𝑖. So, that’s 𝑓 of two plus three 𝑖 over 𝑛. We replace each instance of 𝑥 in our function 𝑓 of 𝑥 with two plus three 𝑖 over 𝑛.
And when we do, we find that 𝑓 of 𝑥 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 𝑥 𝑖 times △𝑥, 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.
