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Lecture

# 6.1 area.pdf

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University of Toronto St. George

Mathematics

MAT237Y1

Dan Dolderman

Fall

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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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