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

# Question Video: Finding the Area under a Curve Using Sigma Notation with Riemann Sums Mathematics • Higher Education

Represent the area under the curve of the function π(π₯) = π₯Β² β 1 on the interval [0, 3] in sigma notation using a right Riemann sum with π subintervals.

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### Video Transcript

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

Remember when weβre finding a right Riemann sum, we find the sum of π₯π₯ times π of π₯ π for values of π from one to π. π₯π₯ is π minus π over π where π and π are the lower and upper limits of our interval, respectively, and π is the number of subintervals. π₯ π is π plus π lots of π₯π₯. We always begin by working out what π₯π₯ is. In our case, π is equal to zero, π is equal to three, and, well, π is just π. This means π₯π₯ is three minus zero over π or just three over π. Next, weβre going to work out what π₯ π is. Itβs π, which we know to be zero, plus π₯π₯, which is three over π, times π. Weβll write this as three π over π.

Of course, we want to know what π of π₯ π is. So it follows that to find π of π₯ π, we find π of three π over π. Letβs substitute three π over π into our formula. Thatβs three π over π all squared minus one which is nine π squared over π squared minus one. Weβre now ready to use the summation formula. Weβre evaluating our sum for values of π from one to π. Its π₯π₯, which is three over π, multiplied by nine π squared over π squared minus one. We distribute our parentheses and then weβ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 π squared. That gives us three π squared over π cubed, leaving us just to simply combine the numerators. We have 27 π squared minus three π squared over π 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 π cubed. Both three and π cubed are indepnent of π. This means we can take three over π cubed outside of the sigma symbol, and that means weβre done. Weβve represented the area under the curve of the function in sigma notation with a right Riemann sum. Itβs three over π cubed times the sum of nine π squared minus π squared for values of π from one π.

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