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Mathematics

MAT237Y1

Dan Dolderman

Fall

Description

Calc 1
Tues 12/2/08
7.1 - Area Between Curves
Remember how we de▯ned the integral? The area under a curve. Well, what if we take it to the
next level and ask ourselves, how do we ▯nd the area between two curves? Think about it for a
second. Use the fact that the integral is the area under the curve, so what it comes down to is if
you want the area between two curves, ▯rst ▯gure out which function is bigger, because that guy
is going to have a larger area (remember that theorem i mentioned where if f(x) ▯ g(x) then
R R
f(x) dx ▯ g(x) dx?). Say, for argument’s sake that f(x) is bigger than g(x) on the interval
we’re dealing with. Well, then ▯nd the area under f(x) and subtract the area under g(x). This
will give you the area between the two curves! Yay!
Notice that it doesn’t matter if, say, f(x) is positive and g(x) is negative, because since g(x) will
end up giving a negative number, when you subtract it you’ll end up adding the area to what you
had before (which kind of makes sense!). So, what does this look like?
Z Z
f(x) dx ▯ g(x) dx
But remember that if you subtract two integrals, you can just combine the two functions under
one integral! Z
f(x) ▯ g(x) dx
HOWEVER, keep in mind that f(x) MUST be greater than or equal to g(x). If not, you’ll get out
the opposite sign that you want! Also, be careful that f(x) and g(x) may switch on you! Think
about the graph of f(x) = x and g(x) = x over the interval [0;2]. g(x) = x is bigger than
2 2
f(x) = x over [0;1], but from [1;2], the two functions switch and x is bigger than x.
This now brings me to explain what i mean when i ask \▯nd the area enclosed between these two
functions." If that’s what I ask, and i say nothing more about the interval, then it is your job to
▯nd out where the two functions intersect (there should be two of them, although it’s possible
there could be more). This is because in between these intersections is where we have an enclosed
area between f and g. So, to do this, set f(x) = g(x) and solve for x. Then, you must ▯gure out
which function is bigger (that’s not too bad...just plug in a point between the two intersections and
see which is bigger. If there were more than two intersections, you need a test point in between
each of them). Once that’s done, then integrate it between to two intersecting points! All right,
enough of this abstract mumbo jumbo. Let’s do a couple examples so that you see what i mean.
2
Example: Find the area enclosed between y = x and y = x.
2
So, ▯rst we need to set these guys equal to each other. x = x. Subtract from both sides and we
get x ▯ x = 0. Now factor x(x ▯ 1) = 0, and we get x = 0 and x = 1 are our intersecting points.
Plug in a test value now to see which is bigger. Let’s do x = 1=2 (that’s between 0 and 1). y = x
gives y = 1=2 and y = x gives y = 1=4. So, y = x is bigger. Therefore, our integral looks like this:
Z
1
x ▯ x dx
0
And now all that’s left is solving this guy
x2 x3▯1 1 1 1
= ▯ ▯ = ▯ ▯ (0 ▯ 0) =
2 3 0 2 3 6
1 2 2
Example: ▯nd the area enclosed by y = 2x ▯ x and y = x .
2 2 2
Where do they intersect? 2x ▯ x = x , so 2x ▯ 2x = 0, which gives us 2(x)(x ▯ 1) = 0 and our
points of intersection are at x = 0 and x = 1. Plug in that test value, say x = 1=2 again.
2 2
2(1=2) ▯ (1=2) = 1 ▯ 1=4 = 3=4. And (1=2) = 1=4. So the ▯rst guy is bigger, which brings us to
Z Z ▯
1 2 2 1 2 2 2 3▯1 2 1
2x ▯ x ▯ x dx = 2x ▯ 2x dx = x ▯ x ▯ = 1 ▯ ▯ (0 ▯ 0) =
0 0 3 0 3 3
Example: ▯nd the area enclosed by y = sin(x) and y = cos(x) between x = 0 and x = ▯=2.
po we must ▯rst ask ourselves when does sin(x) = cos(x)? That special angle of ▯=4 (both equal
2=2). Checking which is bigger (or just thinking about the graph of each of them), we have
cos(x) ▯ sin(x) between [0;▯=4]. However, after that point, sin(x) ▯ cos(x) between [▯=4;▯=2].
This means our area becomes
Z ▯=4 Z ▯=2
cos(x) ▯ sin(x) dx + sin(x) ▯ cos(x) dx
0

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