text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
)dv =
f (t)dt = . . .
a
a
This freedom is very useful, particularly when one or both of the limits of integration depend
on some parameter. In this case, by convention, the dummy variable is chosen to be a different
parameter.
a
a
�
x
�
x
f (x)dx INCORRECT,
f (t)dt CORRECT
a
a
44
18.01 Calculus
Jason Star... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= −1 + cos(2θ).
Notice, by the doubleangle formula for cosine, this equals,
−1 + cos(2θ) = −
2 sin
2(θ).
The hardest step (hidden here) was the geometric computation of f (θ). However, this is completely
unnecessary. Introduce a function,
G(t) =
�
t
√
0
1 − x2dx.
Using symmetry through the yaxis, f (θ) equal... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
area, it has a simple expression using the Riemann integral,
Geometric area =
� b
a
|
f (x)|dx.
Example. Find both the algebraic area and the geometric area bounded by the xaxis and the
graph of y = sin(x) over the interval −π < x < π.
Because sin(x) is an odd function, the area below the xaxis for −π < x < 0 e... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Determine the following Riemann integral to within ±10−4 ,
a
a
�
0.1
0
�
1 +
sin(x)dx.
�
sin(x) has no simple antiderivative. The value of the Riemann integral could be
The expression
approximated well by a Riemann sum. An alternative approach is to use the estimates,
√
√
(1 − x 2
/6)
x ≤
sin(x) ≤
x,
�
for sma... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Theorem of Calculus, the most useful integral
rule is the change of variables rule. The rule for Riemann integrals is nearly the same as the rule
for antiderivatives. The additional feature for Riemann integrals is the change of the limits of
integration.
�
x=b
f (u(x))u�(x)dx =
u=u(b)
f (u)du.
�
x=a
u=u(a)
47
1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
�
π/3
π/4
tan(x)dx =
ln(2)
/2.
It is only fair to note there is a second method. Make the same substitution to simplify the
antiderivative of tan(x) to − ln(|
u|) + C, and then backsubstitute to get,
�
tan(x)dx = − ln( cos(x) ) + C.
|
|
Now use the Fundamental Theorem of Calculus with the original limits of in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ordinary differential equation is an equation involving
a single independent variable x together with a dependent variable y and its derivatives dk y/dxk ,
�
G x, y,
dy d2y
,
dx2
dx
, . . . ,
�
dk y
dxk
= 0.
The largest k for which dk y/dxk occurs in the equation is called order of the differential equation.
E... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ak (x)
dk y
dxk
+ ak−1(x)
dk−1y
dxk−1
+ · · ·
+ a1(x) + a0(x)y = b(x),
dy
dx
for functions ak (x), . . . , a0(x), b(x). If b(x) is zero, the equation is homogeneous. Otherwise it
is inhomogeneous. Very important is the case when all the functions ak (x), . . . , a0(x), b(x) are
constant. Then the differential eq... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.
Thus, the equation y = f (x) is a solution to this separable differential equation.
The algorithm for solving a separable differential equation is the following.
(i). Factor f (x, y) as g(x)/h(y). This is often the most difficult step. In the example, it is quite
easy. Simply take g(x) = −x and h(y) = y.
(ii). Rewrite t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 + C.
C = 1.
(v). Simplify the answer. Often it is best to solve for y = f (x). Often this is unnecessary. It
depends on the problem. In the example problem, the simplest answer is the implicit answer,
So the solution of the initial value problem is,
x2 + y2
= 2
C.
x2 + y2 = 2.
Thus every curve satisfying the g... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
dy/y2/3 = dx. If y equals 0, this equation involves division
by zero. Division by zero is not allowed, so the method breaks down.
Important fact. This fact will not be used in this class. However, it is often crucial in realworld
applications to know the solution to an initial value problem is unique. The fact is, ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
N
dt
= kN.
� dN/N = kdt,
1/N dN = kdt,
ln(|N ) = kt + C.
|
�
N (t) =
N0e . kt
Observe that N (t) increases without bound as t increases. When N is very large, the ecosystem
cannot support such a population. Thus the model is only valid if N (t) is not too large.
A slightly more realistic model hypothesizes a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
rate proportional
to the remaining radioactive isotope. Thus the mass m(t) satisfies a differential equation,
Using the method, the solution is,
dm
dt
= −km.
m(t) = m0e−kt .
53
18.01 Calculus
Jason Starr
Fall 2005
An important feature in decay problems is the halflife. The halflife is the length of time nec... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Tamb.
Lecture 18. October 25, 2005
Homework. Problem Set 5 Part I: (c).
Practice Problems. Course Reader: 3G1, 3G2, 3G4, 3G5.
1. Approximating Riemann integrals. Often, there is no simpler expression for the antideriva
tive than the expression given by the Fundamental Theorem of Calculus. In such cases, the simple... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Δxk /2 is the area of the trapezoid containing the points
(xk−1, 0), (xk−1, yk−1), (xk , 0) and (xk , yk ). In particular, if the graph of y = f (x) is a line, this
trapezoid is precisely the region between the graph and the xaxis over the interval [xk−1, xk ]. Thus,
the approximation above gives the exact integral... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
graph over the pair of intervals by the area under the
unique parabola containing the 3 points (x2l−2, y2l−2), (x2l−1, y2l−1), (x2l, y2l). For notation’s sake,
denote 2l − 1 by k. Thus the 3 points are (xk−1, yk−1), (xk , yk ), and (xk+1, yk+1) (this is slightly
more symmetric).
The first problem is to find the equat... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, yk ) and (xk+1, yk+1) is,
y = A(x − xk )2 + B(x − xk )2 + yk ,
x)2 ,
yk−1 − 2yk + yk+1)/
A
yk+1 − yk−1)/
= (
2(Δ
2(Δ
x).
= (
B
The next problem is to compute the area under the parabola from x = xk−1 to x = xk+1. This is a
straightforward application of the Fundamental Theorem of Calculus,
�
xk+1
xk−1
A(x ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ISimpson =
Writing out the sum gives,
m
b − a �
6m
l=1
(y2l−2 + 4y2l−1 + y2l).
ISimpson =
b−a
6m
((y0 + 4y1 + y2) + (y2 + 4y3 + y4) + (y4 + 4y5 + y6)+
· · · + (y2m−4 + 4y2m−3 + y2m−2) + (y2m−2 + 4y2m−1 + y2m)).
Gathering like terms, ISimpson reduces to,
(b − a)(y0
+ 4 + 2 + 4 + 2 + 4 + 2
y3
y5
y2
y4
y1
y... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
4
6
4
7
4
8
1171
1680
≈
0.6970
b − a
2n
For Simpson’s Rule, because n equals 4, m equals 2. Thus,
ISimpson =
b − a
6m
(y0 + 4y1 + 2y2 + 4y3 + y4) =
1 4
12 4
4
( + 4 + 2 + 4 + ) =
6
4
8
4
5
4
7
1747
2520
≈
0.6933
57
18.01 Calculus
Jason Starr
Fall 2005
According to a calculator, the true va... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the point moves on the
parabola,
and the Pythagorean theorem,
Also, since the speed is constant,
2
y = x ,
L2 = x + y .
2
2
� �2 � �2
+
.
2 s =
dx
dt
dy
dt
This plays the role of the “known rateofchange” in a typical related rates problem.
It is simplest to relate the dependent variables y and L to x. The... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
� �2 �
2 s =
+ 2x
dx
dt
�2
dx
dt
= (1 + 4x 2)
� �2
dx
dt
.
Since s is known to be 3, and x is known to be 2, this equation can be solved for dx/dt,
� �2
dx
dt
=
32
1 + 4(2)2
=
9
.
17
Since the particle is in the first quadrant and moving away from the origin, dx/dt is positive. So,
at the moment whe... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
. An unknown quantity, for instance area A, depends
on some other quantity, for instance the xcoordinate. The method is to allow the independent
variable x vary “infinitesimally” from x to x + dx and then use geometry or science to deduce the
corresponding variation dA of the unknown quantity. The deduction is usual... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(f (x) − g(x))dx.
�
A =
dA =
�
x=b
f (x) − g(x)dx.
x=a
Of course this is the same answer as in the last paragraph. But for other applied integral problems,
the differential method may be the only method that works.
Example. Find the area bounded by the curve y = x(x2 − 3) and a horizontal tangent line.
The hori... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
−x
4
+
3x2
2
2
�
�
�
+ 2x
�
−1
=
27/4.
60
18.01 Calculus
Jason Starr
Fall 2005
3. Volumes of solids of revolutions: the disk method. If the region in the xyplane bounded
by x = a, x = b, y = f (x) and the xaxis is revolved through xyzspace about the xaxis, what is
the volume of the resulting solid? Th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
π
H 2
u=H
2
u
(−du) = π
R2
H 2
�
−
3u
3
0
�
�
�
�
H
.
Evaluating gives the volume,
V =
2
πR H/
3.
In particular, the antiderivative of u2 is responsible for the denominator 3 in the formula for the
volume.
Example. Find the volume of a sphere of radius R.
The sphere of radius R is the solid of revolut... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the cylinder is dx. And the area is the area A(x) of the slice. Thus the
infinitesimal volume of the cylinder is,
Thus the volume is,
dV = Area × width = A(x)dx.
�
V =
dV =
�
x=b
x=a
A(x)dx.
Example. Find the volume of the “corner” region bounded by the xyplane, the xzplane, the
yzplane, and the plane conta... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Calculus
Jason Starr
Fall 2005
5. Volumes of solids of revolution: the washer method. A variation on the disk method
is the washer method. A washer is the solid obtained by removing from a larger disk a concentric
smaller disk of the same width. If the outer radius of the washer is ro and the inner radius is ri,
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Ri
H − x
H
+
Ri + Ro x
.
2 H
Similarly, the equation for the outer radius decreases linearly from ro = Ro at x = 0 to ri =
(Ri + Ro)/2 at x = H. Thus,
ro(x) = Ro
H − x
H
+
Ri + Ro x
.
2 H
2
Since ro − r2
i is a difference of squares, it equals,
2
ro
− r
2
i = (ro + ri)(ro − ri).
This is interesting in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Area × width = π(R2
o − Ri
2)
H − x
H
dx,
giving a total area,
�
V =
dV
=
�
x=H
x=0
π(R2
o − Ri
2)
H − x
H
dx.
Substituting u = H − x, du =
�
−dx gives,
V
=
u=H
π(R2
o − Ri
2)
dx =
u
H
π(R2
2)
o − Ri
2H
�
�
2 H
�
0
u
.
u=0
Thus the total volume of material to produce the dog dish is,
V... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x
∗), the average value is,
k
(b − a)/n.
= f (x
=
∗
k
Average ≈
1
n
n
�
y
∗ .k
k=1
64
18.01 Calculus
Jason Starr
Fall 2005
Multiplying and dividing by Δx gives,
Average ≈
n
1 �
nΔx
k=1
y∗Δx.
k
Now nΔx equals n(a − b)/n, which is a − b. So the average value is,
Average ≈
n
1 �
b − a
k=1
y∗Δx. ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
evaluates to,
1
(2π/ω)
(r0x + (A/ω) sin(ωx)|0
2π/ω
.
1
(2π/ω)
r0(2π/ω) = r0.
Thus, although the radius varies and does not usually equal its ideal value r0, the average value is
indeed,
Average = r0.
2. Average values: nonuniform distribution. It often happens that the average value of f (x) is
desired in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
f (x)ρ(x)dx.
It must be noted, the probability distribution ρ(x) often does not satisfy the normalization condi
tion. In this case, the formula above is wrong. But it is easily correct,
Average =
a f (x)ρ(x)dx)/( � b
( � b
a ρ(x)dx).
Example. A particle is fired through a slit and strikes a screen on the other sid... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= −x2/2σ2 , du = (−x/σ2)dx to reduce this to,
−R
0
� 0
� −R2/2σ2
Ce u(σ2du) +
Ce u(−σ2du) = 2
� 0
Cσ2 e udu.
−R2/2σ2
0
−R2/2σ2
Using the Fundamental Theorem of Calculus, this equals,
2Cσ2 (e −R2/2Σ2 = 2Cσ2(1 − e−R2/2Σ2 ).
u 0
|
As R becomes large, the quantity e−R2/2Σ2 becomes vanishingly small. Thus, in t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1.
The solution is that C = C1/σ. Plugging this into the formula above, the average distance is,
Average distance = 2C1σ,
where,
�
Q
1/C1 = lim
Q→∞ −Q
e−u2/2du.
There is a beautiful argument that,
√
C1 = 1/
2π.
Unfortunately, we cannot yet prove this. Taking it as true gives the final answer,
Average distanc... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
part of a dog dish is a solid of revolution whose
radial crosssection is a triangle of height H whose base has inner radius Ri and outer radius Ro.
Find the volume of material used to make the dog dish.
The volume was computed using the washer method. This time it will be computed using the shell
method. The trian... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x=Ro
V2 =
x=(Ro +Ri)/2
(2πx)
2H
Ro − Ri
(Ro − x)dx =
4πH
Ro − Ri
� x=Ro
x=(Ro+Ri )/2
x(Ro − x)dx.
Believe it or not, this will be simpler to deal with after the substitution u = x − (Ro + Ri)/2,
du = dx. The new integral is
V2 =
4πH
Ro − Ri
�
u=(Ro−Ri)/2
(u + (Ro + Ri)/2)(−u + (Ro − Ri)/2)du.
u=0
Noti... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
0
69
18.01 Calculus
Jason Starr
Fall 2005
By the Fundamental Theorem of Calculus, this equals,
4πH
Ro − Ri
(Ro + Ri)
�
−u
2
2
+
�
(Ro−Ri)/2
�
(Ro − Ri)u
�
�
2
0
.
This evaluates to,
This simplifies to give,
4πH
Ro − Ri
(Ro + Ri)
(Ro − Ri)2
.
8
V = πH(Ro − Ri)(Ro + Ri)/2 = π(R2
o − R2
i )
2.H/
T... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
is a third important way to specify a curve: using parametric equations. Given a parameter
t varying in an interval a ≤ t ≤ b and given functions f (t) and g(t) on this interval, the associated
parametric curve,
�
x = f (t),
y = g(t)
is simply the set of all pairs (x, y) = (f (t), g(t)) as t varies over the interv... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.
This is a parametric equation where time t is the parameter. Even when some other quantity is
the parameter, it is often useful to think of the parameter as time. Thus the curve is the trail left
by a point, or perhaps better, the tip of a pen, as it moves in the plane.
2. Implicitization. Under reasonable hypoth... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2/(2v2
0 cos2(α
)) + tan(
α)(x − x0
) +
y0.
In going from a parametric equation to an implicit equation, there are 2 important warnings to
keep in mind:
71
18.01 Calculus
Jason Starr
Fall 2005
• A parametric equation may traverse only part of the implicit curve. The most usual reason
is that the parameter t i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
of subintervals with endpoints,
a = t0 < t1 < · · · < tn−1 < tn = b.
Approximate the curve on each subinterval tk−1 ≤ t ≤ tk by a line segment. The line segment runs
from the point,
to the point,
(xk−1, yk−1) = (x(tk−1), y(tk−1)),
(xk , yk ) = (x(tk ), y(tk )).
The rise and run of the line segment are,
Δxk = xk... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
� �2 � �2
dx
dt
dy
dt
ds
dt
+
=
,
(−r sin(θ))2 + (r cos(θ))2 = r 2 sin2(θ) + r cos (θ) = r (sin2(θ) + cos 2(θ)) = r .
2
2
2
2
Taking square roots gives the equation,
ds
dθ
= r, ⇔ ds = rdθ.
Thus the arc length of the arc of the circle from θ = a to θ = b is,
�
s =
ds =
� θ=b
θ=a
rdθ =
r(b − a).
This is,... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.
t=a
Since the parameter t in the Riemann integral is only a dummy variable anyway, it is allowed to
replace it by the variable x (so long as x plays no other role in the integral, which it does not).
This gives the formula for the arc length of an explicit curve,
Arc length =
Example 3. Consider the explicit cu... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
/
4 + (1 2) ln(
b/a).
Lecture 22. November 4, 2005
Homework. Problem Set 6 Part I: (f)–(h); Part II: Problem 2 (a) and (c).
Practice Problems. Course Reader: 4G1, 4G4, 4G6, 4H1, 4H3.
1. Surface area of a right circular cone. Before attacking the general problem of the surface
area of a surface of revolution, c... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
75
18.01 Calculus
Jason Starr
Fall 2005
and the average radius of the band is the average of R1 and R2,
By similar triangles,
Rearranging gives,
r = (R1 + R2).
1
2
S2 =
S1
.
R1 R2
R2S1 = R1S2.
Using the formula above, the area of the large cone is,
and the area of the small cone is,
The area A of the ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
as a
line segment. The surface obtained by revolving a line segment is precisely a band of a cone. The
average radius of the cone r is x(t). The slant height of the cone is ds. Thus the area of the band
is,
dA = 2
πrds
= 2
πx
(t)
�� dx
�2 + � dy
dt
dt
�2
dt.
76
18.01 Calculus
Jason Starr
Fall 2005
Integ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
yaxis gives the sphere of radius R. Thus the surface area of the surface of
revolution is the surface area of the sphere of radius R.
As computed in the previous lecture, the differential element of arc length is,
Thus the differential element of surface area is,
ds = Rdθ.
dA = 2πrds = 2πx(θ)(Rdθ) = 2π(R cos(θ))(Rd... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)(cos 2(t) + sin2(t)).
2
The square root is,
�
� �2 � �2 �
dx
dt
+
dy
dt
= 9a2 sin2(t) cos2(t) = 3a sin(t) cos(t).
So the differential element of arc length is,
ds = 3a sin(t) cos(t)dt.
Thus the differential element of surface area of the surface of revolution is,
dA = 2πrds = 2πx(t)ds = 2π(a cos (t))(3a sin(t)... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
r(θ) sin(θ)
a ≤ θ ≤ b.
For each point on the curve, the distance of the point from the origin is,
Distance from origin =
�
x2 + y2 =
√
r2 = r =
| |
�
+r, r ≥ 0,
−r,
r < 0
Also, assuming the point does not equal the origin, the angle of the ray from the origin to the point
is,
Angle = tan−1(y/x) = tan−1(tan(... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Multiplying both sides by r gives,
r = 2a cos(θ),
r 2 = 2ar cos(θ) ⇔ x + y 2 = 2ax.
2
Simplifying this gives the equation,
(x − a)2 + y2 = a2 .
This is the equation of the circle of radius a centered at (a, 0).
4. Sketching polar curves. Given a polar curve, how are we to sketch it? For definiteness,
consider th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
π/4. The curve “goes negative” when
π/4 < θ < 3π/4 and when 5π/4 < θ < 7π/4.
3. Find the extremal values of |r|. A local maximum of r is either a point where r is positive and
a local maximum or a point where r is negative and a local minimum. Similarly for local minima
of |r|. Typically, local maxima of r occur ei... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) = −∞,
θ→a
In our example, r(θ) never limits to ±∞. Thus there are no asymptotes. But in Example B.,
r = a/ sin(θ), r tends to +∞ as θ tends to 0 from above and as θ tends to π from below. Thus
there is an asymptote parallel to the xaxis. Since the explicit equation is y = a, which is a line
parallel to the xax... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the textbook (the sketch was also given in lecture).
Lecture 23. November 8, 2005
Homework. Problem Set 6 Part I: (i) and (j); Part II: Problem 2.
Practice Problems. Course Reader: 4I1, 4I4, 4I6.
1. Tangent lines to parametric curves. This short section was not explicitly discussed for
general parametric curves... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)
2. Tangent lines for polar curves. Although the formula above is perfectly correct, it is a bit
long to remember. There is a slightly different packaging that is much easier to remember. Define
α to be the angle from the horizontal ray emanating from (x(θ), y(θ)) in the positive xdirection,
and the tangent line. T... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
but the formal manipulation leads to the
correct equation.
To compute the denominator in the expression, differentiate both sides of,
to get,
or equivalently,
2
2
2
r = x + y ,
2rdr = 2xdx + 2ydy,
xdx + ydy = r(θ)r�(θ)dθ.
To compute the numerator in the expression, differentiate both sides of,
tan(θ) =
y
,
x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(cos2(θ/2) − sin2(θ/2))
.
−2 sin(θ/2) cos(θ/2)
Replacing 1 − sin2(θ/2) in the numerator by cos2(θ/2), this simplfies to,
2 cos2(θ/2)
−2 sin(θ/2) cos(θ/2)
= − cot(θ/2).
Of course there is an identity,
− cot(u) = tan(u − π/2).
Altogether, this gives,
Therefore,
tan(ψ) = − cot(θ/2) = tan(θ/2 − π/2).
Since α equ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 = [(r�)2 + r ](dθ)2 .
2
Taking square roots gives the differential element of arc length for a polar curve,
ds = �
[r�(θ)]2 + [r(θ)]2dθ.
Example. For the cardioid,
the derivative is,
Thus,
r(θ) = a(1 + cos(θ)),
r�(θ) = −a sin(θ).
(r�)2 + r = a (1 + cos(θ))2 + (−a sin(θ))2 = a (1 + 2 cos(θ) + cos 2(θ)) + a 2 s... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
in a similar way. This was only briefly
discussed in lecture. Here is a continuation of the previous problem.
Example. The top half of the cardioid,
r(θ) = a(1 + cos(θ)), 0 ≤ θ ≤ π,
is revolved about the xaxis to give a fairly good approximation of the surface of an apple. What
is the surface area of this apple?
85 ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
�
u=0
u=1
u (−2du) = 32πa 2
4
�
u=1
u=0
u 4du = 32πa 2
�
5u
5
This evaluates to give the total surface area of the apple,
A = 32
πa
2/5.
1
�
�
�
�
0
.
5. Area of a region enclosed by a polar curve. What is the area of the planar region enclosed
by a cardioid? By the same sort of reasoning as for volumes ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)
�
�
2π
0
.
Evaluating gives,
A =
πa
3
2/2.
Lecture 24. November 15, 2005
Practice Problems. Course Reader: 5A1, 5A2, 5A3, 5A5, 5A6.
1. Inverse functions. Let a, b, s and t be constants. Let y = f (x) be a function defined on the
interval,
and whose values are in the interval,
a ≤ x ≤ b,
s ≤ y ≤ t.
Doe... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
onetoone, because to every output, there is
at most one input. This is the first necessary condition: every invertible function is onetoone.
Next, for every number y in [s, t], there is a number x in [a, b] such that y = f (x). In fact, just
take x to be g(y); then f (x) equals f (g(y)), which equals y. A functio... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
−1(x) is simply to reflect the graph of f (x) through the 45◦
line y = x.
3. The inverse trigonometric functions. The function sin(x) is onetoone and onto on
[−π/2, π/2], taking values in [−1, 1]. Thus there is an inverse function sin−1(x) defined on the
interval [−1, 1], taking values in [−π/2, π/2]. The graph of ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
df (u) = f �(u)du.
88
18.01 Calculus
Jason Starr
Fall 2005
If f has an inverse function g(x), let u be g(x). Then this gives,
df (g(x)) = f �(g(x))dg(x).
On the other hand, f (g(x)) equals x. This gives the formula,
Solving for dg/dx gives,
dx = f �(g(x))dg(x).
d
dx
(g(x)) =
1/f �(g(x)).
This is the f... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
5
This looks remarkably similar to the previous formula. In particular, this gives,
d
dx
(sin−1(x) + cos−1
(x)) =
√
1
1 − x2
+
√
−1
1 − x2
= 0.
Therefore the sum is a constant function. Checking at x = 0 gives the value of this constant
function,
sin−1(x
) + cos
−1(x
) = π/
2.
Finally, because the derivative ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
one part is in
the halfplane where x > 0, and the other part is in the halfplane where x < 0. Because of the
intermediate value theorem, a continuous function x = f (t) cannot jump from x > 0 to x < 0 or
vice versa without crossing x = 0. Thus, refine the question: Are there continuous functions for
the part of th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
these, most of the terms cancel,
cosh2(t) − sinh2(t) =
1
4
((2) − (−2)) = = 1.
4
4
This proves that the parametric curve,
�
x = cosh(t),
sinh(t)
y =
is contained in the righthalf of the hyperbola x2 − y2 = 1. We will see next time that there is an
inverse function of sinh(t), from which it follows that ever... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
+ t) = cosh(s) cosh(t) + sinh(s) sinh(t).
These imply the doubleangle formulas,
sinh(2t) = 2 sinh(t) cosh(t),
cosh(2t) = cosh2(t) + sinh2(t) = 2 cosh2(t) − 1 = 2 sinh2(t) + 1.
From these follow the analogues of the halfangle formulas,
sinh2(t/2) =
1
2
(cosh(t) − 1),
cosh2(t/2) =
1
2
(cosh(t) + 1).
A beautif... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1).
The same type of argument also gives,
cosh−1(x) =
√
ln(x +
x2 − 1), x ≥ 1,
and
tanh
−1(x
) =
(1/
2) ln((1 +
x)/(1 − x)), −1
< x <
1.
2. Derivatives of the inverse hyperbolic functions. By the same methods used to compute
the derivatives of inverse trigonometric functions, the derivatives of the inverse h... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a sin(θ))2 = a (1 − sin2(θ)) = a cos (θ).
2
2
2
2
2
2
93
18.01 Calculus
Jason Starr
Fall 2005
Thus we have,
�
√
a2 − x2dx,
�
x = a sin(θ),
dx = a cos(θ)dθ
, ⇒
�
�
a2 cos2(θ)(a cos(θ)dθ) = a 2
�
2
cos (θ)dθ.
Using the halfangle formula, this becomes,
�
2 a
1
2
+ cos(2θ)dθ = a 2
1
2
�
θ
2
+
1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
types,
u = x +
B
2A
, du = dx,
β2 u + α2, β2 u − α2 , −β2 u + α2 ,
2
2
2
where,
�
|A|, α =
β =
�
B2 − 4AC
|
|
4A|
|
.
Defining a = α/β, finally the integral is transformed to one of 3 possible types,
�
Type I:
√
√
FI (u,
GI (u,
a2 − u2)
a2 − u2)
du,
94
18.01 Calculus
and
�
�
Type II:
Type III: ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
angle formula,
This is easily seen to be,
�
2 a
1
2
−
1
2
cos(2θ)dθ.
�
θ
2
−
1
4
2 a
�
sin(2θ) + C.
Using the doubleangle formula and backsubstituting,
�
√
x2
a2 − x2
dx = (1/
2)(
a2 sin−2(x/a) − x
√
a2 − x2
) +
C.
Alternatively, the hyperbolic inverse substitution is,
x = a tanh(t), dx = asech2... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, this is not a very efficient approach.
Finally, the rational inverse substitution is,
x = a
, dx = a
2t
1 + t2
2(1 − t2)
(1 + t2)2
dt.
The point is that,
Thus the new antiderivative is,
2
2
a − x = a
2 (1 − t2)2
.
(1 + t2)2
�
4a2t2
(1 + t2)2
1 + t2
a(1 − t2)
2a(1 − t2)
(1 + t2)2
dt.
This simplifies ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
a
3
sec (θ)dθ = a 2
1
cos3(θ)
dθ.
This can be simplifed by multiplying numerator and denominator by cos(θ) and then expressing in
terms of sin(θ) as much as possible,
�
2 a
1
cos4(θ)
�
cos(θ)dθ = a 2
1
(1 − sin2(θ))2
cos(θ)dθ.
Make the substitution u = sin(θ), du = cos(θ)dθ) to get,
�
2 a
1
(1 − u2)2 ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a2
) +
C.
Finally, the rational substitution is,
x = a
1 + t2
,
2t
dx = a
−(1 − t2)
2t2
dt.
The point is that,
2
2
a − x = a
2 (1 − t2)2
.
(2t)2
Thus the new antiderivative is,
�
a2(1 + t2)2
4t2
2t −a(1 − t2)
dt.
a(1 − t2)
2t2
This simplifies to,
This evaluates to,
�2
a
−
4
(1 + t2)2
t3
dt = ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
√
√
F (x,
G(x,
Ax2 + Bx + C)
Ax2 + Bx + C)
dx.
For definiteness, consider the example,
�
√
2x
x2 − 2ax + 2a2
dx,
where a is a constant.
Step 1. Complete the square. Complete the square of the expression Ax2 + Bx + C, inside the
radical. In the example,
x − 2ax + 2a 2 = (x − a)2 + a .
2
2
Step 2. Make a line... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
2
2
2
2
With this substitution, the new antiderivative is,
�
a2 tan2(θ) + 2a2 tan(θ) + a
2
�
a2 sec2(θ)
a sec 2(θ)dθ.
99
18.01 Calculus
Jason Starr
Fall 2005
This simplifies to,
�
2 a
(tan 2(θ) + 2 tan(θ) + 1) sec(θ)dθ.
This can be written as a sum of 3 terms,
�
2
a
2
tan (θ) sec(θ)dθ + 2a 2
�
sec(θ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
aster than antidifferentiation, it is usually quite easy to check an antiderivative is correct.
Example. The tricky part is, of course, Step 4. In the example, the integral broke into 3 terms,
�
2
a
2
tan (θ) sec(θ)dθ + 2a 2
�
sec(θ) tan(θ)dθ + a 2
�
sec(θ)dθ.
The last antiderivative was actually Problem 3(b) f... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
sin2(θ)
(1 − sin2(θ))2
cos(θ)dθ.
Now substitute for sin(θ),
The new antiderivative is,
z = sin(θ), dz = cos(θ)dθ.
�
2z
(1 − z2)2
dz.
How do we compute this antiderivative? That is the topic of partial fractions.
Remark: In lecture the solution was done a bit differently. This led to a slightly different an
tid... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
for instance, when
studying Laplace transforms in 18.03. Both polynomials and partial fractions are (relatively) easy
to antidifferentiate. The antiderivative of a polynomial is,
�
q(x)dx =
an
+1)
(n
xn+1 + an−1
n xn + + a1
· · ·
2 x2 + a0x + C.
101
18.01 Calculus
Jason Starr
Fall 2005
The antiderivative o... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Many rational expressions that
come up are not of the simple kinds above. The goal is to express an arbitrary rational expression
as the sum of a polynomial and partial fractions. The first step is polynomial division. Given a
fraction F (x)/G(x), apply polynomial division to get a factorization with remainder,
F (x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
. . . , ak , α1, . . . , αl, and β1, . . . , βl are real numbers. The last l factors were not discussed in
lecture until the end of lecture. Although they are important, they do not often come up in this
course.
The Fundamental Theorem of Algebra asserts that every polynomial with real coefficients has a
factorizatio... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
q(x) is a polynomial, the degree of r(x) is less than the degree of H(x), and r(x) has no
common factor with H(x). This can be further simplified using partial fraction decomposition. It
is a fact that every rational expression r(x)/H(x) can be written in the form,
�
C1,1
x − a1
+
C1,2
(x − a1)2
+ · · ·
+
C1,m... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2005
One approach, which always works but is quite inefficient, is simply to multiply all terms by the
denominator H(x), and then gather coefficients of powers of x. This will give a collection of linear
equations in the unknowns Ci,j , Di,j and Ei,j . There is a unique solution of this set of linear
equations. Methods... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
5.
mining many of the coefficients Ci,j . For each highest power of a linear factor occuring in H(x),
say (x − ai)mi , coverup that term, and substitute x = ai in the remaining polynomial. Then Ci,mi
equals the value,
Ci,mi =
r(x)
H(x)/(x − ai)mi
�
�
�
�
x=ai
.
The proof is quite simple. Multiply every term in... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
z=+1
=
(+1)2
(+2)2
=
1
.
4
Thus the partial fraction decomposition is,
z2
(1 − z2)2
= 1
4
1
+1)
2 + 1
4
(z
1
(z−1)2 + C1,1
z+1 + C2,1
z−1 .
As this example illustrates, the Heaviside coverup method does not always determine all coeffi
cients. However, it reduces the number of coefficients. To find the r... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)(1
/(z
2
+ 1) + 1
/(z − 1)2 − 1/(z
+ 1) + 1
/(z −
1)).
105
18.01 Calculus
Jason Starr
Fall 2005
Using the partial fraction decomposition, the antiderivative is,
�
2
z
(1 −
z2)2
dz =
�
1 −1
4 z + 1
+
−1
z − 1
+ ln( z − 1 / z + 1 ) +C =
| |
|
|
�
�
1
2
z
1 − z2
√
+ ln(
1 − z2) − ln(1 + z) +C.
�
Th... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
dx.
vdu,
�
The new integral is easy to evaluate. Altogether this gives,
�
x cos(x)dx = x sin(x
) + cos( ) +
x
C.
106
18.01 Calculus
Jason Starr
Fall 2005
Because it is much easier to differentiate than the antidifferentiate, it is a good idea to check you
answer.
�
2. How to use integration by parts. The go... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
approach to this problem, which ultimately comes down
to integration by parts again, is to make an inverse substitution,
The new integral is,
x = e , dx = e tdt.
t
�
ln(x)dx =
�
tetdt.
Set u = t and du = etdt. Then u, v, du and dv are,
u = t,
du = dt,
dv = etdt
v = et
107
18.01 Calculus
Using integrati... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
integral is simpler than the original integral. And repeated application of the formula
eventually leads to a formula for the integral. Thus this is a reduction formula. For instance, this
gives,
�
�
[ln(x)]2dx = x[ln(x)]2 − 2
ln(x)dx.
The new integral was already computed. Altogether this gives,
�
[ln(x)]2dx ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
v = − cos(x).
Using integration by parts,
�
�
udv = uv −
�
[sin(x)]ndx = −[sin(x)]n−1 cos(x) + (n − 1)
[sin(x)]n−2 cos (x)dx.
2
vdu,
�
At first blush, this is more complicated than the original integral since it involves both sin(x) and
cos(x). But cos2(x) equals 1 − sin2(x). This substitution gives,
�
[sin... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
F (x) leads to an indeterminate form, it often happens
→
that a more careful computation using calculus eliminates the indeterminate form.
Example. Let b be any real number. Compute the limit as x approaches 0 of F (x) = (b+1/x)−1/x,
x = 0. If we evaluate this limit in a naive manner, we get,
�
lim F (x) = lim b +... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
c2 strictly between a and b such that,
f �(c1)
g�(c2)
=
(f (b) − f (a))/(b − a)
(g(b) − g(a))/(b − a)
=
f (b) − f (a)
.
g(b) − g(a)
Is there a single value c = c1 = c2 where this equality holds?
The answer is yes. Form the function
F (x) = (f (b) − f (a))(g(x) − g(a)) − (g(b) − g(a))(f (x) − f (a)).
Since f (... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a+ g(x) − g(a)
exists if and only if the righthanded limit,
f �(x)
lim
x a+ g�(x)
→
,
exists. If both limits exist, they are equal,
f (x) − f (a)
lim
x→a+ g(x) − g(a)
f �(x)
= lim
x a+ g�(x)
→
.
A similar result holds for lefthanded limits. The proof follows by applying the Generalized Mean
Value Theorem t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
Examples.
limx→a+ f (x)/g(x
) = lim
x→a+ f �(x)/g�(x).
sinh(x)
lim
x→0 sin(x)
= lim
cosh(x)
x→0 cos(x)
1
= =
1
1.
4x3 − 32
lim
x→2 x2 − x − 2
1 − cos(x)
x2
lim
→
x 0
= lim
12x2
x→2 2x − 1
=
12 · 4
2 · 2 − 1
=
48
3
=
16.
= lim sin(x)2x = lim cos(x)2 =
1/2.
x 0
→
x 0
→
4. L’Hospital’s rule ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, and
f �(x)
lim
x a+ g�(x)
→
are defined and nonzero, the formula above can be rewritten as,
Jason Starr
Fall 2005
�
Solving gives,
f (x) −1 �
�
=
lim
x a+ g(x)
→
f �(x)
lim
x→a+ g�(x)
� �
·
�
f (x) −2
lim
x a+ g(x)
→
.
limx→a+ f (x)/g(x
) = lim
x→a+ f �(x)/g�(x),
if both limits are defined and nonzero... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
and (e); Part II: Problems 1 and 2.
Practice Problems. Course Reader: 6B7.
1. A problem with Riemann integrals. Riemann integrals are defined in very many cases.
The result we use most often is that for a piecewise continuous function f (x) on a bounded interval
[a, b], the Riemann integral,
� b
f (x)dx,
exists (a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
∞
a
f (x)dx,
�
∞
a
f (x)dx = lim
t→∞
�
t
a
f (x)dx.
Please note, this is a new definition. It is not a theorem about Riemann integrals.
Example. Let p > 1 be a real number. Then for every t > 1, the integral,
�
t 1
xp
1
dx,
exists and equals,
�
�
�
1
�
�
(p − 1)xp−1
t
1
−
=
1
p − 1
−
1
(p − 1)tp... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x)dx,
is not defined.
3. Improper integrals of the second kind. Here is a second problem with the Riemann
integral. Let [a, b] be a bounded interval. Let f (x) be a function that is bounded on [t, b] for every
a < t < b, but which is unbounded on [a, b]. According to the definition of the Riemann integral,
is not d... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
defined. However, for every 0 < t < 1, the Riemann integral,
is defined equals,
Since 0 < p < 1, the limit,
exists and equals 0. Therefore,
� 1 1
xp
t
dx,
1 − t1−p
.
1 − p
lim t1−p,
0→
t
� 1 1
xp
lim
t→
t
0
dx,
exists and equals 1/(1 − p). Therefore the improper integral,
exists and its value is,
� 1 1... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
→
and if F (x) is monotone decreasing and bounded above, then limx a+ F (x) exists.
→
→
→
These give the following tests for convergence of an improper integral.
Squeezing lemma. If f (x) ≤ g(x) ≤ h(x) on the interval [a, ∞), and if the improper integrals,
�
∞
a
f (x)dx and
�
∞
a
h(x)dx,
exist and are equal, the... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.