Video Transcript
Represent the area under the curve
of the function 𝑓 of 𝑥 equals 𝑥 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 𝛥𝑥 times 𝑓 of 𝑥 𝑖 for values of 𝑖 from one to
𝑛. 𝛥𝑥 is 𝑏 minus 𝑎 over 𝑛 where
𝑎 and 𝑏 are the lower and upper limits of our interval, respectively, and 𝑛 is
the number of subintervals. 𝑥 𝑖 is 𝑎 plus 𝑖 lots of
𝛥𝑥. We always begin by working out what
𝛥𝑥 is. In our case, 𝑎 is equal to zero,
𝑏 is equal to three, and, well, 𝑛 is just 𝑛. This means 𝛥𝑥 is three minus zero
over 𝑛 or just three over 𝑛. Next, we’re going to work out what
𝑥 𝑖 is. It’s 𝑎, which we know to be zero,
plus 𝛥𝑥, 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 𝑥 𝑖 is. So it follows that to find 𝑓 of 𝑥
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 𝛥𝑥, 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.
