Download Monday, April 6: More on Antiderivatives

Monday, April 6: More on Antiderivatives
Math 142, Section 01, Spring 2009
We continue with our study of indefinite integrals and antiderivatives.
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Indefinite Integral Formulas
R
R
f (x) ± g(x) dx =
fR(x) dx ± g(x) dx
k · f (x) dx
= k · f (x) dx k is any constant
1 dx
= x+C
1
xn+1 + C, provided n 6= −1
xn dx
= n+1
x−1 dx
= ln |x| + C
ex dx
= ex + C
sin x dx
= − cos x + C
cos x dx
= sin x + C
2
sec x dx
= tan x + C
sec x tan x dx
= sec x + C
1
√
dx
= arcsin x + C
1−x2
1
dx
= arctan x + C
1+x2
x
b dx
= ln1b bx + C, b > 0 and b 6= 1
You should check each of these formulas. Remember that to check an indefinite integral,
we need only take the derivative. For instance,
(sec x)0 = sec x · tan x,
R
which verifies the formula sec x tan x dx = sec x + C.
In words, the first formula says that the indefinite integral of a sum or difference is the
sum or difference of the indefinite integrals, whereas the second formula says that constants
can be taken out of integrals. The third says the integral of 1 is x. This is actually the
n = 0 case of the fourth rule, but it is worth mentioning it separately.
R
1
− 3ex dx.
Example 1.
(a.) Find √1−x
2
(b.) Is there an antiderivative, F (x), of
1
1+x2
− 3ex satisfying F (0) = 2?
Solution:
For part a we just use our integration formulas:
Z
Z
Z
1
1
√
√
− 3ex dx =
dx − 3 ex dx = arcsin x − 3ex + C.
1 − x2
1 − x2
You should check this by computing its derivative.
For part b : We know from part a that F (x) = arcsin x − 3ex + C is an antiderivative of
√ 1
− 3ex for each constant C. But we want F (0) = 2, so
1−x2
2 = F (0) = arcsin 0 − 3e0 + C = 0 − 3 · 1 + C = C − 3
and so C = 5. Thus, F (x) = arcsin x − 3ex + 5 is an antiderivative of
F (0) = 2.
1
√ 1
1−x2
− 3ex satisfying
2
Application of antiderivatives to Physics (optional)
This section is optional, meaning that you are not responsible for this material. It is
included to illustrate how antiderivatives are often applied in the natural sciences.
Here we derive the equation of motion of an object in free fall near the Earth’s surface.
This equation will be familiar to anyone who has studied physics. What may be unfamiliar
is the approach we follow. We will show this equation of motion is a consequnce of two
more fundamental laws of Physics.
Suppose an object has an initial velocity of v0 and an initial height of h0 . Let t denote
time and h(t) the height of the object at time t. Let m denote the mass of this object.
Newton’s Law of Gravitation: If two objects, with masses m and M , are separated by
a distance of R then the gravitational force, Fg of the one object on the other is
−G · m · M
Fg =
,
R2
where G is the “universal gravitational constant,” which G = 6.673 · 10−11 Newton square
meters per square kilogram.
In our problem, the second object is the Earth, which has a mass of 5.98 · 1024 kg. R is
the separation between or object and the center of the Earth, R = 6.37 · 106 meters. The
gravitational force on our object is therefore
Fg =
−6.673 · 10−11 · 5.98 · 1024 · m
= −9.8 · m.
(6.37 · 106 )2
(We often call g = −9.8 the “gravitational constant;” this is a bit of a misnomer, as the
force Fg is not actually constant. As the object falls, the distance between the object and
the center of the Earth decreases. However, adding a few meters, or even a kilometer to
the radius of the Earth is essentially insignifacant, barely changing the numerical value of
Fg , so assuming the force is constant is fairly harmless.)
Newton’s Second Law Motion: An object of mass m, undergoing a force F , experiences
an acceleration a satisfying F = ma. In our problem the only force acting on the object is
the gravitational force, so
m · a = −9.8 · m.
0
Cancelling the m s tells us a = −9.8. This already tells us something interesting; the mass of
an object does not affect the acceleration of a falling body. (This is neglecting air resistance.
Typically lighter objects do fall more slowly, but that is due to the fact that lighter objects
are more susceptible to air resistance.)
What does this have to do with calculus, and more specifically, antiderivatives and integrals? We want to find a formula for h(t), the height of the object at time t. Now, dh
dt = v(t)
is the velocity of the ball, and
d2 h
dv
= a;
=
dt2
dt
the acceleration is the derivative of the velocity with respect to time, and the second
derivative of the position with respect to time. But a = −9.8 so
dv
= −9.8;
dt
3
taking the antiderivative of v we obtain v(t) = −9.8t + C for some C. But we let v0 denote
the initial velocity, so v(0) = v0 . Thus
v0 = v(0) = −9.8 · 0 + C,
so C = v0 , hence
v(t) = −9.8 · t + v0 .
But we wanted h(t), not v(t), so we take another antiderivative:
dh
= −9.8t + v0
dt
so
1
h(t) = −9.8 · t2 + v0 · t + C = −4.9t2 + v0 · +C
2
for some constant C. But h(0) = h0 , so plugging in t = 0 shows that C = h0 . Therefore
h(t) = −4.9t2 + v0 · t + h0 .
The argument we gave applies rather generally in physics. The basic quantity of interest
is the position, s(t) of an object. Using Newton’s second law and the fact that s00 (t) is the
acceleration we obtain
F
s00 (t) = .
m
Assuming F is nice enough, we can find s(t) by antidifferentiating twice. In practice,
the force often depends on t, s(t), and s0 (t) in rather complicated ways, making it nearly
impossible to explicitly find the antiderivatives. Even in our example above, had we not
made the assumption that Fg was constant we would have been led to
d2 h
−G · M
=
2
dt
(R + h)2
a rather complicated “second-order non-linear ordinary differential equation.” We could
safely approximate the right hand side as a constant since we were assuming that h was
signifacantly smaller than the radius of the Earth, R. This assumption is not always valid.
For instance, in rocket motion the whole point is to have h(t) start at 0 and to end at some
G·M
value significantly larger than the radius of the Earth. Thus, (R+h)
2 ≈ constant is no longer
a reasonable assumption.