Video: Finding the Area under a Curve Using Sigma Notation with Riemann Sums

Represent the area under the curve of the function 𝑓(π‘₯) = π‘₯Β² βˆ’ 1 on the interval [0, 3] in sigma notation using a right Riemann sum with 𝑛 subintervals.

02:17

Video Transcript

Represent the area under the curve of the function 𝑓 of π‘₯ equals π‘₯ squared minus one on the close interval zero to three in sigma notation using a right Riemann sum with 𝑛 subintervals.

Remember when we’re finding a right Riemann sum, we find the sum of π›₯π‘₯ times 𝑓 of π‘₯ 𝑖 for values of 𝑖 from one to 𝑛. π›₯π‘₯ is 𝑏 minus π‘Ž over 𝑛 where π‘Ž and 𝑏 are the lower and upper limits of our interval, respectively, and 𝑛 is the number of subintervals. π‘₯ 𝑖 is π‘Ž plus 𝑖 lots of π›₯π‘₯. We always begin by working out what π›₯π‘₯ is. In our case, π‘Ž is equal to zero, 𝑏 is equal to three, and, well, 𝑛 is just 𝑛. This means π›₯π‘₯ is three minus zero over 𝑛 or just three over 𝑛. Next, we’re going to work out what π‘₯ 𝑖 is. It’s π‘Ž, which we know to be zero, plus π›₯π‘₯, which is three over 𝑛, times 𝑖. We’ll write this as three 𝑖 over 𝑛.

Of course, we want to know what 𝑓 of π‘₯ 𝑖 is. So it follows that to find 𝑓 of π‘₯ 𝑖, we find 𝑓 of three 𝑖 over 𝑛. Let’s substitute three 𝑖 over 𝑛 into our formula. That’s three 𝑖 over 𝑛 all squared minus one which is nine 𝑖 squared over 𝑛 squared minus one. We’re now ready to use the summation formula. We’re evaluating our sum for values of 𝑖 from one to 𝑛. Its π›₯π‘₯, which is three over 𝑛, multiplied by nine 𝑖 squared over 𝑛 squared minus one. We distribute our parentheses and then we’re going to look to create a common denominator. We can do that by multiplying both the numerator and denominator of our second fraction by 𝑛 squared. That gives us three 𝑛 squared over 𝑛 cubed, leaving us just to simply combine the numerators. We have 27 𝑖 squared minus three 𝑛 squared over 𝑛 cubed.

Now, we can simplify this somewhat. The numerators share a factor of 27 and three. And of course they have a common denominator of 𝑛 cubed. Both three and 𝑛 cubed are indepnent of 𝑖. This means we can take three over 𝑛 cubed outside of the sigma symbol, and that means we’re done. We’ve represented the area under the curve of the function in sigma notation with a right Riemann sum. It’s three over 𝑛 cubed times the sum of nine 𝑖 squared minus 𝑛 squared for values of 𝑖 from one 𝑛.

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