Area
- Page ID
- 219418
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Area Area of a Rectangle and Using Rectangles to Approximate Area Under a Curve Recall that the area of a rectangle is the height times the base. What if we wanted to paint a wall that has a ceiling the shape of y = x2 , a flat floor and a right wall at x = 2 yards and a left wall at x = 5 yards.
We can approximate the area by cutting out 6 rectangles. Since the base of the wall is 5 - 2 yards long, and there are 6 rectangles, the base of each rectangle is (5 - 2)/6 = .5 yards. The height of each rectangle is the y-coordinate of the left side of each rectangle. The x- coordinates are 2 + 0(.5), 2 + 1(.5), 2 + 2(.5), 2 + 3(.5), 2 + 4(.5), 2 + 5(.5) (2 + 0(.5))2 ,(2 + 1(.5))2 , (2 + 2(.5))2 , (2 + 3(.5))2 , (2 + 4(.5))2 , (2 + 5(.5))2 We see that the ith rectangle has y coordinate: height = (2 + i(.5))2 = 4 + 2i + .25i2 To get the area of the ith rectangle we multiply the height by the base: (4 + 2i + .25i2)(.5) Finally to get the total area we add the terms up: S[(4 + 2i + .25i2)(.5)] This will be a lower bound for the area. Exercise:
Left and Right Sums If we take the limit as i approaches infinity, We arrive at the formulae:
Usually to compute a definite integral, we use left or right sums. Example Use the right sum to find Solution: \( \lim\limits_{n \to \infty} \sum_{i=0}^{n} [ 2(1+\frac{2i}{n}) + 1] \frac{2}{n} = \lim\limits_{n \to \infty} \sum_{i=0}^{n} [ 3+\frac{4i}{n})] \frac{2}{n} \) \( = \lim\limits_{n \to \infty} \sum_{i=0}^{n} [ \frac{6}{n}] + \frac{8}{n^2}\sum_{i=0}^{n} i = \lim\limits_{n \to \infty} (6 + \frac{8}{n^2}\frac{n(n+1)}{2}) \) \( = \lim\limits_{n \to \infty} (6 + \frac{8n^2 + 8n}{2n^2}) = 6 + 4 = 10\)
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