Video: Approximating the Area under the Curve of a Function Using Riemann Sums

Use a right Riemann sum to approximate the area under the curve of 𝑓(π‘₯) = π‘₯Β² βˆ’ 4 in the interval [2, 4]. Use subintervals with 𝑛 = 5 .

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Video Transcript

Use a right Riemann sum to approximate the area under the curve of 𝑓 of π‘₯ equals π‘₯ squared minus four in the closed interval two to four. Use subintervals with 𝑛 equals five.

Remember, a right Riemann sum is a way of approximating the area between a curve and the π‘₯-axis by splitting it into rectangles. In a right Riemann sum, the height of each rectangle is the value of the function at the right-end point of the subinterval. Let’s sketch out the graph of our function in our closed interval to see what this might look like.

The graph of 𝑦 equals π‘₯ squared minus four is this parabola shape. It intersects the π‘₯-axis at π‘₯ equals two. We’re looking to find this region, the region between the curve and the π‘₯-axis. And the first thing that we’re going to do is split it up into 𝑛 subintervals, where 𝑛 is equal to five.

Now, we can kind of work this out on our head, but there is a formula. This says that the width of each subinterval Ξ”π‘₯ is equal to 𝑏 minus π‘Ž over 𝑛, where π‘₯ equals π‘Ž and π‘₯ equals 𝑏 are the vertical lines that bound our region and 𝑛 is the number of subintervals. In this case, 𝑏 is equal to four, π‘Ž is equal to two, and we’re told that 𝑛 is equal to five. So, the width of each subinterval is two-fifths, or 0.4.

We’re going to split our region up then into subintervals each with a width of 0.4. So, starting at π‘₯ equals two, we get 2.4, 2.8, 3.2, 3.6, and then we end up at four. This is a right Riemann sum. So, the height of each rectangle is the value of the function at the right end of each subinterval. And so, we can add in our rectangles as shown.

Now, an approximation for the area between the curve 𝑓 of π‘₯ equals π‘₯ squared minus four and the π‘₯-axis is the area of the first rectangle plus the area of the second rectangle plus the area of the third rectangle plus the area of the fourth rectangle plus the area of the fifth rectangle. Now, each of these rectangles sits above the π‘₯-axis, so we shouldn’t be expecting any negative values. But if they lay below the π‘₯-axis, we would be looking to subtract the absolute value for the area.

The area of our first rectangle is its width multiplied by its height. Well, we know its width now to be 0.4 and its height is the value of the function at 2.4. 𝑓 of 2.4 is 2.4 squared minus four, which is 1.76. And so, π‘Ž one is 0.4 times 1.76. We then have π‘Ž two. That’s 0.4 times the value of the function π‘₯ equals 2.8. This time, 𝑓 of 2.8 is 2.8 squared minus four, which is equal to 3.84. And so, π‘Ž two is 0.4 times 3.84. π‘Ž three is 0.4 times 𝑓 of 3.2. π‘Ž four is 0.4 times 𝑓 of 3.6. And π‘Ž five is 0.4 times the value of the function when π‘₯ is equal to four.

𝑓 of 3.2 is 6.24. 𝑓 of 3.6 is 8.96. And 𝑓 of four is four squared minus four, which is 12. And so, our right Riemann sum is 0.4 times 1.76 plus 0.4 times 3.84 plus 0.4 times 6.24 plus 0.4 times 8.96 plus 0.4 times 12. That gives us 328 over 25. And so, an approximation to the area under the curve of 𝑓 of π‘₯ equals π‘₯ squared minus four in the interval two to four is 328 over 25 square units.

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