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Lecture

# 6.1 area.pdf

3 Pages
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School
University of Toronto St. George
Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Wed 11/5/08 6.1 - Area Finally, we go on to chapter 6, which studies the other major problem in calculus: area. ...particularly, area beneath curves. So, how do we ▯nd the area between a non-negative function f(x) and the x-axis (which we call the area beneath f(x)) in some interval [a;b]? If you think about it, in general, this sounds like a pretty di▯cult topic. It might not be too bad if the function forms certain geometric shapes that we know area formulas for, but how are we able to calculate this stu▯ in general? Fortunately, for now, we are going to only focus on the ones that form nice geometric shapes, and then later we’ll move on to ▯nding the area for the other functions. Example: Take a look at the line f(x) = 2x from [0;2]. When you draw it on a graph, it basically looks like a triangle (draw the line x = 2 to see the triangle better). What’s the formula for the area of a triangle?1bh, where in this case, b = 2 and h = 4 (just plug in f(2) = 4...this came from [0;2], our 2 1 interval). So, the area underneath f(x) = 2x is in f2ct▯ 4 = 4. Remember, we’re only interested in the area between the x-axis and the function itself. Whenever you are given a function that’s a line (ie, f(x) is a polynomial where the highest power of x is 1), there are four geometric ▯gures that you might encounter that you should know how to calculate the area for. One is a triangle, which was just demonstrated in the above example. A second geometric ▯gure is a rectangle (or square), the third is a trapezoid, and the fourth would be a semicircle. Remember what the area of a trapezoid is? Well, we have the two parallel sides b and b and 1 1 2 then the height, h. The area turns out to be: A = (1 + b2)h 2 Example: Find the area underneath f(x) = 3x ▯ 3 over [2;4]. Solution: Again, graph this. Try to see why f(x) = 3x ▯ 3 starting at 2 and ending at 4 gives a trapezoid. Our two parallel sides will be at the lines x = 2 and x = 4. Each of the parallel sides has length 3(2) ▯ 3 = 6 ▯ 3 = 3 and 3(4) ▯ 3 = 12 ▯ 3 = 9. Combining that with h = 4 ▯ 2 = 2 (the distance between the parallel sides), we get A = 1(3 + 9)(2) = 12(2)= 12 2 2 One more example should su▯ce to show that we can combine more than one geometric ▯gure in one function. Example: ▯nd the area underneath 8 <2x x < 3 f(x) = > 6 3 ▯ x < 6 : ▯6x + 42 x ▯ 6 in the interval [1;7]. Solution: First graph this guy. It’s a piecewise function and we can actually split it up into di▯erent geometric ▯gures. Namely, we get a trapezoid between 1 and 3 and then a rectangle 1 between 3 and 6 and ▯nally a triangle between 6 and 7. So, just treat it like three separate problems! Trapezoid: 1 8(2) A = 2[2(1) + 2(3)][(3 ▯ 1)] = 2 = 8 Rectangle: A = 6(6 ▯ 3) = 6(3) = 18 Triangle: 1 6 A = (7 ▯ 6)(▯6(6) + 42) = = 3 2 2 Adding these all together, we get 8 + 18 + 3 = 29. So, if we have a function f(x), call the area function underneath f(x) (between a and x) A(x). The really neat thing is that this area function A(x) is not just something created out of the blue because it’s fun to think about. A(x) actually relates to all that derivative stu▯ that we have done for weeks now. This relation is: A (x) = f(x). Whoa. Let’s think about this with an example. Example: Let f(x) = 4. Find A(x) over the interval [▯2;x]. Solution: Now wait a minute! That interval in this example is a bit weird! Well, it’s exactly what’s above in how A(x) is
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