The Friday Wrap, Sept 3
Monday: Fundamental theorem of calculus. Nutshell version: the derivative
of the integral is the original function. Formal version:
Z x
d
f (t) dt = f (x).
dx a
Fine print: f has to be continuous for this to work.
Souped up version:
d
dx
Z
b(x)
a(x)
f (t) dt = a′ (x)f (a(x)) − b′ (x)f (b(x)).
Application to definite integrals version: If, somehow or other, you’ve got an
antiderivative F for f , that is, a function F so that F ′ (x) = f (x), then
b
Z
f (t) dt = F (b) − F (a).
a
Now some examples, some worked, some left to you as recommended homework.
Examples:
1.
d
dx
2.
Z
x
0
d
dx
3.
d
dx
Z
1
1
dt =
.
1 + t4
1 + x4
Z
π
x2
sin(t)
dt = ?
t
2x
x
t3 dt = 2(2x)3 − 1(x3 ) = 7x3 .
(This one is easy to check).
4. Let
(
1, if x > 0;
s(x) =
−1 if x < 0.
Rx
Let F (x) = 0 s(t) dt. Then F (x) = |x| and while it is differentiable at
positive values of x, where the derivative is 1 = s(x), and also differentiable
at negative values of x, where the derivative is −1 and thus again equal
to s(x), F ′ (0) does not exist, and since there’s no such thing as F ′ (0), in
particular it’s not true that F ′ (0) = 0 = s(0).
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Wednesday: Areas between curves. Nutshell: the area is the integral of the
length of a cross-sectional line. Formal version: Area of the region bounded by
an upper curve u(x) and a lower curve l(x) that intersect at x = a and at x = b,
with b > a, is
Z b
(u(x) − l(x)) dx.
A=
x=a
Twists on the basic idea: the integration can also be done with respect to y,
and then, it’s
Z
A = right boundary(y) − left(y) dy.
Two other complications: finding a or b may be difficult, and it may happen
that the curves intersect each other somewhere between a and b, so that in
(a, c), u(x) > l(x), while in (c, b), it’s the other way around.
Examples:
1. Find the area bounded between the curves y = x and y = x3 . Careful,
the answer is not zero.
R1
2. Find 0 sin−1 (x) dx. Hint: Use the integrating sideways idea.
Friday: Volumes. Nutshell version: a volume is computed by integrating the
area of cross sections. Formal version: if a solid is bounded below by the plane
Z = a and above by the plane Z = b, and if the area of the cross section in the
plane Z = z is A(z), then the volume of the solid is given by
V =
Z
b
A(z) dz.
z=a
When the solid is a solid of revolution about the z axis got by revolving some
reasonably simple shape in the x, y plane about the z axis, the cross sectional
area A(z) takes the form A(z) = π(R2 (z) − r2 (z)) dz where R(z) is the outer
radius and r(z) is the inner radius.
Other shapes have to be dealt with on a case by case basis and some ingenuity, either via good visualization or by way of algebra, will be required.
Examples:
1. A cone of height H and base radius R. Consider this as a solid of revolution
got by revolving the triangle with corners (x = 0, z = 0), (x = 0, z = H),
and (x = R, z = H) about the z axis. There is no inner radius to worry
about, and the outer radius r(z) is given by r = Rz/H because of similar
triangles. Thus the volume is
Z H
π(Rz/H)2 dz = πR2 H/3.
V =
z=0
2. Revolve the shape bounded by y = x and y = x2 about the x axis. The
volume works out to 2π/15; this is example 4 from the text, page 433.
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3. Revolve the same shape about the y axis. This is example 6, p 435.
4. Revolve the same shape about the line y = −1, a line parallel to the x-axis.
The example of two intersecting cylinders of the same shape is difficult but
instructive.
= z are squares
√ The cross sections in planes of the form Z √
√ of sidelength 2 1 − z 2 , because when Z = z, x ranges from − 1 − z 2 to 1 − z 2 and
so does y. Integrating 4(1 − z 2 ) from z = −1 to z = 1 gives the volume.
The example of a cheese slice bounded by the planes z = 0 and z = x out of
the solid cylinder x2 +y 2 ≤ 1 led √
to cross sections in√the vertical plane
√X = x be2 , 0), (x, − 1 − x2 , x),(x, + 1 − x2 , x),
1
−
x
ing rectangles
with
corners
(x,
−
√
√
and (x, + 1 − x2 , 0). This rectangle has area 2x 1 − x2 so that
Z
1
x=0
p
2x 1 − x2 dx
gives the volume.
The example of the volume of a tent with a square floor and diagonal curved
poles can be given a variety of details to flesh it out. Here’s one. Assume
the poles run from (−2, −2, 0) through (0, 0, 2) to (2, 2, 0) and from (−2, 2, 0)
through (0, 0, 2) to (2, −2, 0) and, when bent, take the shape of a parabola.
This parabola would then have to be something like z = Ax2 = Ay 2 , measuring
z down from the center point of the roof of the tent and x and y in distance
sideways from that point. From the given corners, 2 = 4A so A = 1/2. Now
the cross sections at distance z below the roof are square, with area (2x)(2y) =
R2
4xy = 4x2 = 8z. Thus, the volume is z=0 8z dz = 4z 2 |2z=0 = 16 (cubic meters).
That’s half the volume of a rectangular solid, 4 by 4 by 2.
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