# 132class3

```The Fundamental Theorem of Calculus, Part 1
If f is continuous on [a, b], then the function
g defined by
Z x
g (x) =
f (t) dt a ≤ x ≤ b
a
is continuous on [a, b] and differentiable on
(a, b), and
g 0(x) = f (x).
The Fundamental Theorem of Calculus, Part 2
If f is continuous on [a, b], then
Z b
f (x) dx = F (b) − F (a),
a
where F (x) is any antiderivative of f (x),
that is, a function such that
F 0(x) = f (x).
Application of FTC
Example Evaluate
R6 1
3 x
dx.
Solution An antiderivative for x1 is
F (x) = ln x. So, by FTC,
Z 6
1
dx = F (6) − F (3) = ln 6 − ln 3.
x
3
Example Evaluate
R3
1
e x dx.
Solution Note that an antiderivative for e x
is F (x) = e x . So, by FTC,
Z 3
e x dx = F (3) − F (1) = e 3 − e.
1
Example Find area A under the cosine curve
from 0 to b, where 0 ≤ b ≤ π2 .
Solution Since an antiderivative of f (x) = cos(x) is
F (x) = sin(x), we have
b
Z b
A=
cos(x) dx = sin(x) = sin(b)−sin(0) = sin(b).
0
0
The Fundamental Theorem of Calculus
Suppose f is continuous on [a, b].
Rx
1. If g (x) = a f (t) dt, then
g 0(x) = f (x).
Rb
2. a f (x) dx = F (b) − F (a), where
F (x) is any antiderivative of f (x),
that is, F 0(x) = f (x).
Notation: Indefinite integral
R
f (x) dx = F (x) means F 0(x) = f (x).
R
We use the notation f (x) dx to denote an
antiderivative for f (x) and it is called an
indefinite integral. A definite integral has
the form:
b
Z b
Z
f (x) dx =
f (x) dx = F (b) − F (a)
a
a
Table of Indefinite Integrals
R
R
R
c &middot; f (x) dx =
R c &middot; f (x) dx
R
[f (x) + g (x)]
dx
=
f
(x)
dx
+
g (x) dx
R
k dx = kx + C
R n
n+1
x dxR= xn+1 + C
(n 6= −1)
1
= ln |x| + C
R x dx
e x dx = e x + C
Table of Indefinite Integrals
R
Rsin x dx = − cos x + C
R cos2 x dx = sin x + C
R sec2 x dx = tan x + C
R csc x dx = − cot x + C
R sec x tan x dx = sec x + C
csc
csc x + C
R x1cot x dx = −
−1
x +C
2 +1 dx = tan
x
R 1
−1
√
dx
=
sin
x +C
2
1−x
R2
3
3
2x
−
6x
+
0
x 2 +1 dx and
interpret the result in terms of areas.
Solution FTC gives
Z 2
3
3
2x − 6x + 2
dx
x +1
0
2
x2
x4
−1
= 2 − 6 + 3 tan x .
4
2
0
2
1
= x 4 − 3x 2 + 3 tan−1 x
2
0
1
= (24 ) − 3(22 ) + 3 tan−1 2 − 0
2
= −4 + 3 tan−1 2
Example Find
The Net Change Theorem
The Net Change Theorem The integral
of a rate of change is the net change:
Z b
F 0(x) dx = F (b) − F (a).
a
Example A particle moves along a line so that its
velocity at time t is v (t) = t 2 − t − 6 (measured in
meters per second). Find the displacement of the
particle during the time period 1 ≤ t ≤ 4.
Solution By the Net Change Theorem, the
displacement is
Z 4
Z 4
s(4) − s(1) =
v (t)dt =
(t 2 − t − 6) dt
1
t3 t2
=
− − 6t
3
2
1
4
1
9
=− .
2
This means that the particle moved 4.5 m toward
the left.
Example A particle moves along a line so that its
velocity at time t is v (t) = t 2 − t − 6 (measured in
meters per second). Find the distance traveled
during the time period 1 ≤ t ≤ 4.
Solution Note that v (t) = t 2 − t − 6 = (t − 3)(t + 2) and
so v (t) ≤ 0 on the interval [1, 3] and v (t) ≥ 0 on [3, 4]. From
the Net Change Theorem, the distance traveled is
Z 4
Z 3
Z 4
|v (t)| dt =
[−v (t)] dt +
v (t) dt
1
1
Z
=
1
3
(−t 2 + t + 6) dt +
3
Z
4
(t 2 − t − 6) dt
3
3
3 3
4
t
t2
t
t2
− − 6t
= − + + 6t +
3
2
3
2
1
3
61
=
≈ 10.17m
6
The Substitution Rule
The Substitution Rule is one of the main
tools used in this class for finding
antiderivatives. It comes from the Chain
Rule:
[F (g (x))]0 = F 0(g (x))g 0(x).
So,
Z
F 0(g (x))g 0(x) dx = F (g (x)).
The Substitution Rule
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Find x 3 cos(x 4 + 2) dx.
The Substitution Rule
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Find x 3 cos(x 4 + 2) dx.
Solution
1. Make the substitution: u = x 4 + 2.
3
2. Get
Z du = 4x dx.
Z
Z
1
1
3
4
cos u du
x cos(x + 2) dx = cos u &middot; du =
4
4
1
1
= sin u + C = sin(x 4 + 2) + C .
4
4
Note at the final stage we return to the original variable x.
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
Example Evaluate
R√
2x + 1 dx.
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R√
Example Evaluate
2x + 1 dx.
Solution Let u = 2x + 1. Then du = 2 dx, so dx =
the Substitution Rule gives
Z
Z
Z
√
√ du
1
1
2x + 1 dx =
u
=
u 2 du
2
2
3
du
.
2
3
1 u2
1 3
1
= &middot; 3 + C = u 2 + C = (2x + 1) 2 + C .
2 2
3
3
Thus,
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Calculate e 5x dx.
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Calculate e 5x dx.
Solution If we let u = 5x, then du = 5 dx, so dx = 15 du.
Therefore
Z
Z
1
1
1
5x
e dx =
e u du = e u + C = e 5x + C .
5
5
5
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Calculate tan x dx.
The Substitution Rule If u = g (x) is a differentiable
function whose range is an interval I and f is continuous on I ,
then
Z
Z
0
f (g (x))g (x) dx = f (u) du.
R
Example Calculate tan x dx.
Solution
Z
Z
sin x
tan x dx =
dx.
cos x
This suggests substitution u = cos x, since then
du = − sin x dx and so, sin x dx = −du:
Z
Z
Z
sin x
du
dx = −
tan x dx =
cos x
u
= − ln |u| + C = − ln | cos x| + C .
The Substitution Rule for Definite Integrals
Substitution Rule for Definite Integrals
If g 0 is continuous on [a, b] and f is
continuous on the range of u = g (x), then
Z b
Z g (b)
f (g (x))g 0(x) dx =
f (u) du.
a
g (a)
The Substitution Rule for Definite Integrals
If g 0 is continuous on [a, b] and f is continuous on the range
of u = g (x), then
Z b
Z g (b)
0
f (g (x))g (x) dx =
f (u) du.
a
Example Calculate
g (a)
Re
1
ln x
x
dx.
The Substitution Rule for Definite Integrals
If g 0 is continuous on [a, b] and f is continuous on the range
of u = g (x), then
Z b
Z g (b)
0
f (g (x))g (x) dx =
f (u) du.
a
g (a)
Re
Example Calculate 1 lnxx dx.
Solution We let u = ln x because its differential du = dx
x
occurs in the integral. When x = 1, u = ln 1 = 0; when x = e,
u = ln e = 1. Thus
1
Z 1
Z e
ln x
u2
1
u du =
dx =
= .
x
2 0 2
0
1
Areas between curves
Area between curves
The area A of the region bounded by the curves
y = f (x), y = g (x), and the lines x = a, x = b,
where f and g continuous and f (x) ≥ g (x) for all x
in [a, b], is
Z b
A=
[f (x) − g (x)] dx
a
Example Find the area of the region bounded
above by y = e x , bounded below by y = x, and
bounded on the sides by x = 0 and x = 1.
Example
Find the area of the region bounded above by
y = e , bounded below by y = x, and bounded on the sides
x
by x = 0 and x = 1.
Solution The region is shown in the figure. The upper
boundary curve is y = e x and the lower boundary curve is
y = x. So we use the area formula with f (x) = e x , g (x) = x,
a = 0, and b = 1:
1
Z 1
1 2
x
x
A=
(e − x) dx = e − x
2
0
0
1
= e − − 1 = e − 1.5
2
Example Find the area of the region enclosed by
the parabolas y = x 2 and y = 2x − x 2 .
Example Find the area of the region enclosed by
the parabolas y = x 2 and y = 2x − x 2 .
Solution We first find the points of intersection of the
parabolas by solving their equations simultaneously. This gives
x 2 = 2x − x 2 , or 2x 2 − 2x = 0. Thus 2x(x − 1) = 0, so x = 0
or 1. The points of intersection are (0, 0) and (1, 1). So the
total area is:
h 2
i1
R1
R1
x
x3
2
2
A = 0 (2x − 2x ) dx = 2 0 (x − x ) dx = 2 2 − 3
0
= 2 12 − 31 = 13
Areas between curves
The area between the curves y = f (x) and
y = g (x) and between x = a and x = b is
Z b
A=
|f (x) − g (x)| dx
a
.
Example Find the area enclosed by the line
y = x − 1 and the parabola y 2 = 2x + 6.
Example
Find the area enclosed by the line y = x − 1
and the parabola y 2 = 2x + 6.
Volumes
Volumes
V = volume ≈
Pn
∗
i=1 A(xi )∆x
Definition of volume
Definition of Volume Let S be a solid that lies
between x = a and x = b. If the cross-sectional
area of S in the plane Px , through x and
perpendicular to the x−axis, is A(x), where A is a
continuous function, then the volume of S is
Z b
n
X
V = lim
A(xi∗ )∆x =
A(x) dx
n→∞
i=1
a
Computing volume of a sphere
Computing volume of a sphere
Example Show the volume of a sphere of radius r
is V = 43 πr 3 .
Computing volume of a sphere
Example Show the volume of a sphere of radius r
is V = 43 πr 3 .
Solution x 2 + y 2 = r 2
p
y = r 2 − y 2.
2
So the
is A(x) = πy 2 = π(r
x 2 ). So,
R rcrossectionalRarea
Rr −
r
2
2
2
V = −r A(x) dx = −r π(r − x ) dx = 2π 0 (r − x 2 ) dx
h
ir
3
3
= 2π r 2 x − x3 = 2π r 3 − r3 = 34 πr 3
0
Computing a volume of revolution
Example
Find the volume of the solid obtained by
rotating about the x-axis the region under the curve y =
from 0 to 1.
√
x
Computing a volume of revolution
Example
Find the volume of the solid obtained by
rotating about the x-axis the region under the curve y =
from 0 to 1.
Solution√The area of the crossection is:
A(x) = π( x)2 = πx.
i1
R1
R1
2
So, V = 0 A(x) dx = 0 πx dx = π x2 =
0
π
2
√
x
Volume of a solid paraboloid
Example
Find the volume of the solid obtained by
rotating the region bounded by y = x 3 , y = 8 and x = 0
Volume of a solid paraboloid
Example
Find the volume of the solid obtained by
rotating the region bounded by y = x 3 , y = 8 and x = 0
√
Solution Note that x = 3 y . The area of crossection is:
2
√
A(y ) = πx 2 = π( 3 y )2 = πy 3 . So,
h 5 i8
R∞
R ∞ 2 dy
3
V = 0 A(y ) dy = 0 πy
= π 53 y 3 =
0
96π
.
5
Other volumes
Example
2
The region R enclosed by the curves y = x and
y = x is rotated about the x-axis. Find the volume of the
solid region.
Other volumes
Example
The region R enclosed by the curves y = x and
2
y = x is rotated about the x-axis. Find the volume of the
solid region.
Solution The curves y = x and y = x 2 intersect at the
points (0, 0) and (1, 1). Crossection of rotated surface has the
shape of a washer (annular ring). So the crossectional area is:
A(x) = πx 2 − π(x 2 )2 = π(x 2 − x 4 ).
h 3
i
R1
R1
5
So, V = 0 A(x) dx = 0 π(x 2 − x 2 ) dx = π x3 − x5 = 2π
.
15
Example
The region R enclosed by the curves y = x and
y = x 2 is rotated about the line y = 2. Find the volume.
Example
The region R enclosed by the curves y = x and
y = x 2 is rotated about the line y = 2. Find the volume.
Solution The crossection is again a washer. The
crossectional area is:
A(x) = π(2 − x 2 )2 − π(2 − x)2 = π(x 4 − 5x 2 + 4x).
R1
R1
So, V = 0 A(x) dx = π 0 (x 4 − 5x 2 + 4x) dx =
h 5
i1
3
2
π x5 − 5 x3 + 4 x2 = 8π
.
15
0
Integration by parts
first method was substraction which came from the
chain rule. Integration by parts comes from the
product rule.
d
[f (x)g (x)] = f (x)g 0 (x) + g (x)f 0 (x).
dx
So,
f (x)g 0 (x) = (f (x)g (x))0 − g (x)f 0 (x)
Thus, after taking antiderivatives, we get
Z
Z
f (x)g 0 (x) = f (x)g (x) − g (x)f 0 (x) dx.
Integration by parts
Z
f (x)g 0 (x) dx = f (x)g (x) −
Z
g (x)f 0 (x) dx
The above is called the formula for integration
by parts.
Integration by parts
Z
f (x)g 0 (x) dx = f (x)g (x) −
Z
g (x)f 0 (x) dx
The above is called the formula for integration
by parts.
If we let u = f (x) and v = g (x), then du = f 0 (x) dx
and dv = g 0 (x) dx, so the formula becomes:
Z
Z
u dv = uv − v du.
Strategy for using integration by parts
Recall the integration by parts formula:
Z
Z
u dv = uv − v du.
To apply this formula we must choose dv so that we
can integrate it! Frequently, we choose u so that the
derivative of u is simpler than u. If both properties
hold, then you have made the correct choice.
Examples using strategy
Z
1.
x sin x dx : Choose u = x and dv = sin x dx
Examples using strategy
Z
1.
x sin x dx : Choose u = x and dv = sin x dx
Z
2.
xe x dx : Choose u = x and dv = e x dx
Examples using strategy
Z
1.
x sin x dx : Choose u = x and dv = sin x dx
Z
2.
Z
3.
xe x dx : Choose u = x and dv = e x dx
t 2 e t dt : Choose u = t 2 and dv = e t dt
Examples using strategy
Z
1.
x sin x dx : Choose u = x and dv = sin x dx
Z
2.
Z
3.
4.
Z
xe x dx : Choose u = x and dv = e x dx
t 2 e t dt : Choose u = t 2 and dv = e t dt
x 2 sin 2x dx : Choose u = x 2 and v = sin 2x dx
Examples using strategy
Z
1.
x sin x dx : Choose u = x and dv = sin x dx
Z
2.
Z
3.
4.
Z
xe x dx : Choose u = x and dv = e x dx
t 2 e t dt : Choose u = t 2 and dv = e t dt
x 2 sin 2x dx : Choose u = x 2 and v = sin 2x dx
Z
5.
ln(x) dx : Choose u = ln x and v = dx
```
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