Download ANTIDERIVATIVES AND AREAS AND THINGS 1. Integration is

ANTIDERIVATIVES AND AREAS AND THINGS
ANTHONY CAINE
1. Integration is Generalized Multiplication
What I will refer to as Generalized Multiplication, others may call Generalized Area. However, objects and
applications where we see integrals occur will be more general than finding an area.
Recall: The spring scale example we saw on the first day
For example, I may want you to calculate the area under the curve f (x) =
1
x2
for all 1 < x < 9?
The amazing thing is, it turns out we can. And the amazinger thing is, all we need to know
is a function whose
d
derivative is x12 ! In fact, let’s do it now. Find a function, let’s call it F (x), so that dx
F (x) = x12 = x−2 .
F (x) =
For example, I’ll use g(x) = 2 −
1
x.
,
Your function may look similar but don’t use the same one as me.
Okay, now that we have a function, F (x), whose derivative is x−2 , to find the area under this graph between
x = 1 and x = 9, we simply evaluate F (9) − F (1). For example, with my function
1
g(9) − g(1) = 2 − 9−1 − (2 − 1−1 ) = 1 − .
9
What do you get for your function?
F (9) − F (1) =
.
Wow,
Remark. Now, you have no reason to believe me when I say the area is 1 − 91 , but it is, and we will find out why
later.
What we just used to calculate the area under a curve between two points is a big result called The Fundamental
Theorem of Calculus and understanding this theorem will be one of our main goals for the course. That said, let’s
just go a head and state:
Theorem 1 (The Area Theorem). If f is a function that is continuous on the interval [a, b] ⊆ R and F (x) is ANY
function that satisfies F 0 (x) = f (x), then
Z b
f (x)dx = F (b) − F (a).
a
Remark. The symbol
b”
Rb
a
means “the area under the curve from a to b” or, equivalently, “The integral from a to
For now, let’s “write” The Area Theorem using a picture and words. It says: If f (x) is continuous then
1
ANTIDERIVATIVES AND AREAS AND THINGS
2
Let’s use this to calculate some areas of familiar objects. The first one I will do and the second one you should
try, but please ask if you need help.
Example. Using The Area Theorem, calculate the area of a rectangle with base, b, and height h.
To use The Area Theorem we need to express this rectangle as part of the graph of a function.
So, a good starting point is to draw the rectangle in the xy-plane. It is convenient to draw one corner at the
origin:
Rb
Now, remember, our goal is to find the area of this rectangle, using a f (x) dx. Does it make since that we
should use a =
and b =
(the base length)? Does it make since that f (x) should be the constant
function h, i.e. f (x) = h?
.
.
If this does not make since, then please raise your hand and let’s talk about it.
.
.
Using these values, we can say the area of the rectangle is
Z b
Z b
f (x) dx =
h dx.
0
0
To solve this, we need to think of a function whose derivative is h. That is, we need to find F (x) so that
F 0 (x) = h. What should we use? F (x) =
. Now The Area Theorem says
Z
b
Z
0
b
h dx = F (b) − F (0) =
f (x) dx =
0
and presto, we get the area of a rectangle!
Let’s do another.
Exercise 1. Using The Area Theorem, find the area of a 45-45-90 triangle with side length b. (Such as this one)
ANTIDERIVATIVES AND AREAS AND THINGS
3
.
.
.
.
.
.
Key Idea. Back in middle school or high school, you learned Area = base x height. Integration generalizes this
concept for when the height is changing and says
Z
Area =
height(x) dx
base
that is, the area is the integral of a height function (i.e. a height that depends on your location) over a base. The
above formula is what I will call the Area Concept. Because that’s exactly what it is, it’s just a concept, or an
outline, we will use.
You may be thinking, “Why does this matter?”
Obviously, the first reason is, it’s fun. (Then again, I like math.) The second reason is, areas turn out to
measure valuable properties.
Example. A “spring scale” has one side attached to the wall and I pull the other. Like this:
Let’s pretend it’s a big scale and I can stretch the spring by 1 meter. Then I can graph the amount of force I
apply in relation to the distance of stretch. It’s looks like this:
Now, we might want to figure out how much work we have done stretching this spring. High school physics tells
us, “work = force × distance.” But in our case the force is changing with respect to distance! So, now what? The
answer is to use an integral. Just like how, when the height of an area is changing, area = base × height became
Z
Area =
height(x) dx
base
In this case, work = force × distance becomes
Z
Work =
f orce(x) dx
distance
So I claim the work stretching the spring from x = 0 to x = p is
Z p
kx dx =
0
and if you know your physics, you may recognize this as the work formula for a spring!
.
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.
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ANTIDERIVATIVES AND AREAS AND THINGS
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Problems.
1. (a) What is a function whose derivative is x2 ?
(b) What is a function whose derivative is x3 ?
(c) What is function whose derivative is x6 ?
(d) Suppose n is a positive integer (e.g. 1, 2, 3, ...). What is a function whose derivative is xn ?
2. (a) Using what you found in question 1, and the The Area Theorem, what is the area under the curve xn
R1
between 0 and 1? That is, calculate 0 xn dx.
Z
1
xn dx =
0
R1
(b) Find the limit as n goes to infinity of 0 xn dx.
.
.
.
.
.
.
(c) Does this limit make intuitive sense to you? Why or why not? Hint: if you take a number less than 1 (for
example .9) and raise it to a really big power (for example 100) you get a really small number (.9100 is really tiny).
(See graph)
.
.
.
.
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ANTIDERIVATIVES AND AREAS AND THINGS
5
2. Antiderivatives
Review From Last Time. Last time we stated The Area Theorem, which says: If f is a function that is continuous on the
and F (x) is ANY function that satisfies F 0 (x) = f (x), then
Z
b
f (x)dx = F (b) − F (a)
a
Then, to give support to this theorem, we showed that using The Area Theorem we can get familiar area
formulas, such as that of a square and a triangle. As a refresher, let’s calculate
7
Z
4x3 dx =
3
Z
π
2
cos(x) dx =
0
Antiderivatives.
Definition 1. Given a function f (x), an antiderivative of f (x) is a function whose derivative is f (x).
For example, an antiderivative of f (x) = 4x3 is F (x) = x4 , because F 0 = f . Another possible antiderivative is
G(x) = x4 +1000, because G0 (x) = f (x). Another is (your turn) H(x) =
, because H 0 (x) = f (x).
Exercise 2. Find a possible antiderivative for the following functions:
(1) cos(x) has
as an antiderivative.
(2) sin(x) has
as an antiderivative.
1
x
has
as an antiderivative.
(4) ex has
as an antiderivative.
(3)
(5) For positive n, xn has
as an antiderivative.
Remark. In general, an antiderivative is not unique. For example, for the function f (x) = 4x3 , any function of
d
the form x4 + c, where c is a constant, will be an antiderivative of f (x) = 4x3 , because dx
(x4 + c) = 4x3 = f (x).
However, antiderivatives will have a common “form”
R
Definition 2. Given a function f (x), the antiderivative of f (x), denoted “ f (x) dx” is a function with a parameter
that tells us what all possible antiderivatives of f (x) look like.
Example.
Z
Exercise 3.
Z
cos(x) dx =
4x3 dx = x4 + c here, c is the parameter that can change
Z
,
x
e dx =
Z
1
dx =
x
Z
, for n > 0
xn dx =
Recall, a derivative is linear. This means
d
d
d
(c · f (x) + g(x)) = c ·
(f (x)) +
(g(x)) = c · f 0 (x) + g 0 (x).
dx
dx
dx
It turns out, an antiderivative is too:
Z
Z
Z
(c · f (x) + g(x)) dx = c · f (x) dx + g(x) dx.
ANTIDERIVATIVES AND AREAS AND THINGS
Exercise 4.
Z
4x3 + cos(x) −
ex
dx =
2
6
.
Also, antiderivatives are called ANTI-derivatives because if we are given a derivative of a function we can almost
recover the function:
Z
f 0 (x) dx = f (x) + c, but we have to worry about this pesky c.
Remark. Everything I have said up to this point has basically been me repeating myself in different ways and
with new words. Our first objective in 21B boils down to this:
In order to calculate
R
f (x) dx we need a function whose derivative is f (x).
With this goal in mind, it’s good to understand how derivatives work, so that we may undo them.
2
• Calculate the derivative of f (x) = ex ,
• Calculate the derivative of g(x) = (x2 + 1)−1 ,
• What “rule” did you use to calculate these derivatives?
Now let’s do some integrals of familiar functions:
• Calculate
Z
3
2
2xex dx =
0
• Calculate
Z
3
2x
dx =
+ 1)2
0
R3
If you are like many others, calculating 0
this is for a very good reason, namely:
(x2
2x
(x2 +1)2
dx is a lot scarier/harder than calculating
d
2
−1
dx (x + 1) .
And
Remark. Calculating derivatives is easy because you only need to know the chain rule and the product rule!
However, for an integral or an antiderivative, you need to actually know or figure out a function whose derivative
is the thing you are integrating! (So we need to starting working backwards.)
U-Substitution.
R3
Example. Suppose we want to find 0 2x(x2 − 5)2 dx. We could multiply out the polynomial and integrate term
by term. OR, we can do a u-sub. It works like this:
d
I realize that dx
(x2 − 5) = 2x and it motivates me to define u(x) = x2 − 5. This means du
dx = 2x. Or, solving
1 ˙
for “dx,” we find 2x du = dx. We now substitute these new values into the old integral:
Z 3
Z u(3)
Z 4
1 ˙
43 (−5)3
189
2
2
2
2
2x(x
−
5
)
dx
=
2x(u)
·
du
=
u
du
=
−
=
= 63
| {z }
2x{z }
3
3
3
0
u(0)
−5
|
this is u
the limits change
this was dx
Now let’s do the same integral without a u-sub:
Z 3
Z 3
Z 3
2
2
4
2
2x(x − 5) dx =
2x(x − 10x + 25) dx =
2x5 − 20x3 + 50x dx
0
0
0
3
6
4
6
20x
3
2x
=
−
+ 25x2 = ( − 5 · 34 + 25 · 32 ) − 0
6
4
3
0
= 34 (3 − 5) + 225 = −2(81) + 225 = 225 − 162 = 63
We got the same thing!!!! Indeed, a u-sub when done correctly (i.e. changing limits and being consistent) is a
legal move and you get the same answer as you would if you chose to do a lot of algebra.
Remark. We use u-subs to simplify integrals.
ANTIDERIVATIVES AND AREAS AND THINGS
7
Example. Calculate
Z
4(6x − 1)2/3 dx.
R
Answer: This integral looks scary. Where as x2/3 dx is much more manageable. So, let’s try u(x) = 6x − 1.
1
Then du
dx = 6 and dx = 6 du. Subbing everything in gives
Z
Z
Z
4
2 u5/3
2 (6x − 1)5/3
1
du =
u2/3 du = ·
+c= ·
+ c.
4(6x − 1)2/3 dx = 4 u2/3 ·
6
3 5/3
3
5/3
|6{z }
this was dx
Notice, we put everything in terms of x again in the end, because the problem began in terms of x.
Exercise 5.
Z
(3x − 2)4 dx =
U-substitution can also be thought of as undoing the chain rule and this happens to help us simplify integrals.
d
Recall, the chain rule says dx
(f (g(x))) = f 0 (g(x)) · g 0 (x).
The trick is realizing when a chain rule has occurred inside the integral, and using u(x) = g(x).
Example.
Z
(2x + 1)
p
x2 + x − 1 dx =?
√
Answer: Do you see how x2 + x − 1·(2x+1) is in the form f 0 (g(x))·g 0 (x)I. This time I will use u(x) = x2 +x−1.
1
Why? Because then du/dx = 2x + 1 and 2x+1
du = dx, which will simplify nicely when substituing.
Z
Z
Z
p
√
√
1
2
udu = 2/3 · u3/2 + c = 2/3 · (x2 + x − 1)3/2 + c
(2x + 1) x + x − 1 dx = (2x + 1) u
du =
2x + 1
Exercise 6.
Z
8x
dx =
(4x2 + 1)1/3
Z
cos(x) · sin(x) dx =
Previously I had you calculate
have used a u-sub technique.
Example. Let u(x) = x2 then
1
2x
R3
0
2
2xex dx and
R3
2x
0 (x2 +1)2
du = dx and
Z
3
2xe
x2
Z
and using u(x) =
+ 1 we get dx =
Z
3
0
du
2x
32
dx =
02
0
x2
dx. These are two more examples where we could
9
(eu ) du = eu 0
and so
2x
dx =
2
(x + 1)2
Z
u(3)
u(0)
u−2
10
du = −u−1 .
1
d
(f (g(x))) = f 0 (g(x)) · g 0 (x) is in the integral, and we are setting
Just to reitterate, we are realizing when dx
u(x) = g(x). Analyzing the two examples above,
2
2xėx = eg(x) ġ 0 (x) =
d x2
2x
1
d
(e ), and 2
=
· g 0 (x) =
((g(x))−1 ).
2
2
dx
(x + 1)
(g(x))
dx
IIf you don’t see this please ask. That said, to see it, notice g(x) = x2 + x − 1. I know this because its derivative shows up, i.e.
0
g (x) = 2x + 1 appears. Thus f 0 (x) must be
√
√
x. Hence f 0 (g(x)) = x2 + x − 1
ANTIDERIVATIVES AND AREAS AND THINGS
8
Problems.
(1)
Z 3
x2 (1 − 4x3 )1/2 dx =
1
(2)
(a)
Z
x dx =
(b)
Z
e2x · (2e2x ) dx =
(c) Given a function, f (x), what is
Z
f (x) · f 0 (x) dx =
.HintII
(d) Given two functions, f (x) and g(x),
Z
f 0 (g(x)) · g 0 (x) dx =
.
.
.
.
.
.
.
.
.
IIYour answer will be a function of f (x). If you can’t figure it out, look at what happened in the two previous cases, and the case
R
sin(x)cos(x) dx.