**Unformatted text preview: **Definition (Definition of Derivative)
Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x)
is given by the following formula if the limit exists:
f (x + h)
h!0
h f 0 (x) = lim f (x) Definition (Definition of Derivative)
Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x)
is given by the following formula if the limit exists:
f (x + h)
h!0
h f 0 (x) = lim
Example
f (x) = x 2 f (x) Definition (Definition of Derivative)
Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x)
is given by the following formula if the limit exists:
f (x + h)
h!0
h f (x) f 0 (x) = lim
Example
f (x) = x 2
f (x + h)
h!0
h
lim f (x) = lim h!0 (x + h)2
h x2 = x 2 + 2hx + h2
h x2 Definition (Definition of Derivative)
Let f : R ! R be a real valued function f (x). Then the derivative f 0 (x)
is given by the following formula if the limit exists:
f (x + h)
h!0
h f (x) f 0 (x) = lim
Example
f (x) = x 2
f (x + h)
h!0
h
lim f (x) = lim h!0 = lim h!0 (x + h)2
h x2 = x 2 + 2hx + h2
h 2hx + h2
= 2x + h = 2x
h x2 Powers, Constant and Sum Rules
Let c 2 R and f (x), g (x) be real valued functions. Then:
(x c )0 = cx c 1 . (cf (x))0 = cf 0 (x).
(f (x) + g (x))0 = f 0 (x) + g 0 (x). Powers, Constant and Sum Rules
Let c 2 R and f (x), g (x) be real valued functions. Then:
(x c )0 = cx c 1 . (cf (x))0 = cf 0 (x).
(f (x) + g (x))0 = f 0 (x) + g 0 (x).
Example
If F (x) = 10x 2 + 7x, then
F 0 (x) Powers, Constant and Sum Rules
Let c 2 R and f (x), g (x) be real valued functions. Then:
(x c )0 = cx c 1 . (cf (x))0 = cf 0 (x).
(f (x) + g (x))0 = f 0 (x) + g 0 (x).
Example
If F (x) = 10x 2 + 7x, then
F 0 (x) = 10(2x) + 7 = 20x + 7. Exponential Rule
For f (x) = e x , f 0 (x) = e x Exponential Rule
For f (x) = e x , f 0 (x) = e x Derivatives of sin(x) and cos(x)
sin0 (x) = cos(x)
cos0 (x) = sin(x) Exponential Rule
For f (x) = e x , f 0 (x) = e x Derivatives of sin(x) and cos(x)
sin0 (x) = cos(x)
cos0 (x) = sin(x) Example
If f (x) = 5e x 3 sin(x), find f 0 (x). Exponential Rule
f 0 (x) = e x For f (x) = e x , Derivatives of sin(x) and cos(x)
sin0 (x) = cos(x)
cos0 (x) = sin(x) Example
If f (x) = 5e x 3 sin(x), find f 0 (x). f 0 (x) = (5e x 3 sin(x))0 Exponential Rule
f 0 (x) = e x For f (x) = e x , Derivatives of sin(x) and cos(x)
sin0 (x) = cos(x)
cos0 (x) = sin(x) Example
If f (x) = 5e x 3 sin(x), find f 0 (x). f 0 (x) = (5e x 3 sin(x))0 = (5e x )0 (3 sin(x))0 = 5e x 3 cos(x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2 Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = ex
1+x 2 . Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = ex
1+x 2 . Solution.
f 0 (x) = d
d
(1 + x 2 ) dx
(e x ) e x dx
(1 + x 2 )
(1 + x 2 )2 Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = ex
1+x 2 . Solution.
f 0 (x) =
= d
d
(1 + x 2 ) dx
(e x ) e x dx
(1 + x 2 )
(1 + x 2 )2 (1 + x 2 )e x e x (2x)
(1 + x 2 )2 Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = ex
1+x 2 . Solution.
f 0 (x) =
= d
d
(1 + x 2 ) dx
(e x ) e x dx
(1 + x 2 )
(1 + x 2 )2 (1 + x 2 )e x e x (2x)
e x (1 x)2
=
.
2
2
(1 + x )
(1 + x 2 )2 Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2 Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) = sin(x)
.
cos(x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) =
Solution (using the quotient rule):
✓
◆0
sin(x)
tan0 (x) =
cos(x) sin(x)
.
cos(x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) = sin(x)
.
cos(x) Solution (using the quotient rule):
✓
◆0
sin(x)
cos(x) cos(x) sin(x)( sin(x)
tan0 (x) =
=
cos(x)
cos2 (x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) = sin(x)
.
cos(x) Solution (using the quotient rule):
✓
◆0
sin(x)
cos(x) cos(x) sin(x)( sin(x)
tan0 (x) =
=
cos(x)
cos2 (x)
= cos2 (x) + sin2 (x)
cos2 (x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) = sin(x)
.
cos(x) Solution (using the quotient rule):
✓
◆0
sin(x)
cos(x) cos(x) sin(x)( sin(x)
tan0 (x) =
=
cos(x)
cos2 (x)
= cos2 (x) + sin2 (x)
1
=
2
cos (x)
cos2 (x) Product and quotient rules
(f (x)g (x))0 = f 0 (x)g (x) + f (x)g 0 (x) 0
f (x)
f 0 (x)g (x) f (x)g 0 (x)
=
g (x)
(g (x))2
Example
Compute f 0 (x) for f (x) = tan(x) = sin(x)
.
cos(x) Solution (using the quotient rule):
✓
◆0
sin(x)
cos(x) cos(x) sin(x)( sin(x)
tan0 (x) =
=
cos(x)
cos2 (x)
= cos2 (x) + sin2 (x)
1
=
= sec2 (x).
2
cos (x)
cos2 (x) Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = e 3x x 2 , find f 0 (x). Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = e 3x x 2 , find f 0 (x). f 0 (x) = (e 3x x 2 )0 = (e 3x )0 (x 2 )0 Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = e 3x x 2 , find f 0 (x). f 0 (x) = (e 3x x 2 )0 = (e 3x )0 (x 2 )0 = 3e 3x 2x Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
Find F 0 (x) if F (x) = e x 2 +1 . Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
Find F 0 (x) if F (x) = e x 2 +1 . Solution:
Let g (x) = e x and f (x) = x 2 + 1. Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
Find F 0 (x) if F (x) = e x 2 +1 . Solution:
Let g (x) = e x and f (x) = x 2 + 1.
Then F (x) = (g f )(x). Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
Find F 0 (x) if F (x) = e x 2 +1 . Solution:
Let g (x) = e x and f (x) = x 2 + 1.
Then F (x) = (g
So, f )(x). F 0 (x) = g 0 (f (x)) · f 0 (x) Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
Find F 0 (x) if F (x) = e x 2 +1 . Solution:
Let g (x) = e x and f (x) = x 2 + 1.
Then F (x) = (g
So, f )(x). F 0 (x) = g 0 (f (x)) · f 0 (x) = e x 2 +1 (2x) Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Solution:
Draw the triangle with angle ✓ such that x = tan(✓).
This
side x, adjacent side 1 and hypotenuse
p triangle has opposite
1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 . Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Solution:
Draw the triangle with angle ✓ such that x = tan(✓).
This
side x, adjacent side 1 and hypotenuse
p triangle has opposite
1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 .
Be definition of inverse function, x = tan(tan 1 (x). Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Solution:
Draw the triangle with angle ✓ such that x = tan(✓).
This
side x, adjacent side 1 and hypotenuse
p triangle has opposite
1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 .
Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives:
1 = tan0 (tan 1 (x)) · (tan 1 0 ) (x) Chain Rule
(f g )0 (x) = f 0 (g (x)) · g 0 (x) Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Solution:
Draw the triangle with angle ✓ such that x = tan(✓).
This
side x, adjacent side 1 and hypotenuse
p triangle has opposite
1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 .
Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives:
1 = tan0 (tan
= sec2 (tan 1 (x)) · (tan 1 (x)) · (tan 1 0 ) (x) 1 0 ) (x) Chain Rule
g )0 (x) = f 0 (g (x)) · g 0 (x) (f
Example
If f (x) = tan 1 (x), show f 0 (x) = 1
.
1 + x2 Solution:
Draw the triangle with angle ✓ such that x = tan(✓).
This
side x, adjacent side 1 and hypotenuse
p triangle has opposite
1 + x 2 , and so, sec2 [✓ = tan 1 (x)] = 1 + x 2 .
Be definition of inverse function, x = tan(tan 1 (x). Take the derivative of each side and the chain rule gives:
1 = tan0 (tan
= sec2 (tan 1 (x)) · (tan Solving we get, (tan 1 0 ) (x) = 1 (x)) · (tan 1 0 1 0 ) (x) ) (x) = (1 + x 2 ) · (tan 1
1+x 2 . 1 0 ) (x). Derivatives of classical functions
(x c )0 = cx c 1 (e x )0 = e x
sin0 (x) = cos(x)
cos0 (x) = sin(x) 0 tan (x) = sec2 (x)
ln0 (x) = x1 , where ln(x) is the natural log.
log0a (x) = 1
x ln(a) , 1
d
dx (sin (x))
d
1
dx (tan (x)) =
= where loga (x) is the log in base a. p 1
1 x2
1
1+x 2 Definition (Antiderivatives)
A function F is called an antiderivative
of f on an interval I if
Z
0
F (x) = f (x) for all x in I. We use
f (x) dx to denote F (x). Definition (Antiderivatives)
A function F is called an antiderivative
of f on an interval I if
Z
0
F (x) = f (x) for all x in I. We use
f (x) dx to denote F (x).
Example
Let f (x) = x 2 . Then an antiderivative F (x) for x 2 is F (x) = x3
3 . Definition (Antiderivatives)
A function F is called an antiderivative
of f on an interval I if
Z
0
F (x) = f (x) for all x in I. We use
f (x) dx to denote F (x).
Example
Let f (x) = x 2 . Then an antiderivative F (x) for x 2 is F (x) = x3
3 . Theorem
If F is an antiderivative of f on an interval I, then the most general
antiderivative of f on I is
F (x) + C
where C is an arbitrary constant. Table of Anti-di↵erentiation Formulas
Let F (x), G (x) be the antiderivative respectively for the
functions f (x), g (x).
Function
c · f (x)
f (x) + g (x)
x n (n 6= 1)
1
x
x e
cos x Particular antiderivative
c · F (x)
F (x) + G (x)
x n+1
n+1 ln |x|
ex
sin x Table of Anti-di↵erentiation Formulas
Function
sin x
sec2 x
sec x tan x
p 1
1 x2
1
1+x 2 Particular antiderivative
cos x
tan x
sec x
sin 1 x
tan 1 x Area under y = x 2 from 0 to 1.
Example
Use rectangles to estimate the area under the parabola y = x 2 from 0 to
1. Area estimate using right end points 1
R4 = ·
4
Note area A < 15
32 ✓ ◆2
✓ ◆2
✓ ◆2
1
1
1
1
3
1
15
+ ·
+ ·
+ · 12 =
4
4
2
4
4
4
32
= .46875 Area estimate using left end points L4 = 1 2 1
·0 + ·
4
4 ✓ ◆2
✓ ◆2
✓ ◆2
1
1
1
1
3
7
+ ·
+ ·
=
= .21875
4
4
2
4
4
32 Note area A satisfies
.21875 A .46875 Area definition using right end points
Definition
The area A of the region S that lies under the graph of the continuous
function f is the limit of the sum of the areas of approximating
rectangles:
A = lim Rn
n!1 = lim [f (x1 ) x + f (x2 ) x + · · · + f (xn ) x]
n!1 Note in the above definition that if I = [a, b], then
x= b a
n , where n is the number of rectangles or divisions. Definition of a Definite Integral Definition
If f is a continuous function defined for a x b, we divide the
interval [a, b] into n subintervals of equal width x = (b a)/n.
We let x0 (= a), x1 , x2 , . . . , xn (= b) be the endpoints of these
subintervals and we let x1⇤ , x2⇤ , . . . , xn⇤ be any sample points in
these subintervals, so xi⇤ lies in the i-th subinterval [xi 1,xi ].
Then the definite integral of f from a to b is
Z b f (x) dx = lim
a n!1 X
i=1 f (xi⇤ ) x Figure: Here there are n subintervals with b = xn .
Example (Trapezoidal Approximation)
The best simple estimate of A with n subintervals is the ”trapezoidal
approximation”: A= b a
n f (a)/2 + f (b)/2 + n 1
X
i=1 f (xi ) ! Properties of the Integral
Z b
Z
[f (x) + g (x)] dx =
a Z
Z b c · f (x) dx = c a Z a f (x) dx +
a Z b g (x) dx
a b f (x) dx, where c is any constant
a b [f (x) b g (x)] dx = Z b f (x) dx
a Z b g (x) dx
a Definition
A function F defined on real valued functions is said to be linear if given
two such functions, f , g ,
F (f + g ) = F (f ) + F (g )
For c 2 R, F (c · f ) = c · F (f ).
Question
Why do you think integration is said to be linear? Theorem (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 di↵erentiable on (a, b), and
g 0 (x) = f (x).
Theorem (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 Z 6
3 1
dx.
x Application of FTC Example
Evaluate Z 6
3 1
dx.
x Solution:
An antiderivative for x1 is F (x) = ln x.
So, by FTC,
Z 6
1
dx = F (6) F (3) = ln 6
x
3 ln 3. Example
Evaluate Z 3
1 e x dx. Example
Evaluate Z 3 e x dx. 1 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
1 e. Example
Find area A under the cosine curve from 0 to b, where 0 b ⇡
2. Example
Find area A under the cosine curve from 0 to b, where 0 b ⇡
2. Solution:
Since an antiderivative of f (x) = cos(x) is F (x) = sin(x), we have
Z b
b
A=
cos(x) dx = sin(x) = sin(b) sin(0) = sin(b).
0 0 Theorem (Fundamental Theorem of Calculus)
Suppose f is continuous on [a, b].
Z x
If g (x) =
f (t) dt, then g 0 (x) = f (x).
Z a b f (x) dx = F (b)
a 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:
Z b f (x) dx =
a Z b f (x) dx = F (b)
a F (a) Table of Indefinite Integrals
Z
Z
c · f (x) dx = c · f (x) dx
Z [f (x) + g (x)] dx = Z Z Z f (x) dx + Z g (x) dx k dx = kx + C x n dx =
Z Z x n+1
+C
n+1 (n 6= 1
dx = ln |x| + C
x
e x dx = e x + C 1) 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:
0 [F (g (x))] = F 0 (g (x))g 0 (x).
So, Z F 0 (g (x))g 0 (x) dx = F (g (x)). Substitution Rule: If u = g (x) is a di↵erentiable function whose
range is an interval I and f is continuous on I, then
Z
Z
f (g (x))g 0 (x) dx = f (u) du.
Example
Find Z x 3 cos(x 4 + 2) dx. Substitution Rule: If u = g (x) is a di↵erentiable function whose
range is an interval I and f is continuous on I, then
Z
Z
f (g (x))g 0 (x) dx = f (u) du.
Example
Find Z x 3 cos(x 4 + 2) dx. Solution:
Make the substitution: u = x 4 + 2.
Get du = 4x 3 dx.
Z
Z
Z
1
1
3
4
x cos(x + 2) dx = cos u · du =
cos u 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.
= Example
Evaluate Z p 2x + 1 dx. Example
Evaluate Z p 2x + 1 dx. Solution:
Let u = 2x + 1.
Then du = 2 dx, so dx = du
2 . The Substitution Rule gives
Z
Z
Z
p
p du
1
1
2x + 1 dx =
u
=
u 2 du
2
2
3 3
1 u2
1 3
1
= · 3 + C = u 2 + C = (2x + 1) 2 + C .
2 2
3
3 Substitution Rule: If u = g (x) is a di↵erentiable 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
Calculate Z e 5x dx. Substitution Rule: If u = g (x) is a di↵erentiable 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
Calculate Z e 5x dx. Solution:
If we let u = 5x, then du = 5 dx, so dx =
Therefore
Z e 5x 1
dx =
5 Z 1
5 du 1
1
e u du = e u + C = e 5x + C .
5
5 Substitution Rule: If u = g (x) is a di↵erentiable 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
Calculate R tan x dx. Substitution Rule: If u = g (x) is a di↵erentiable 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
Calculate
Solution: R tan x dx.
Z tan x dx = Z sin x
dx.
cos x This suggests substitution u = cos x, since then du =
and so, sin x dx = du:
Z
Z
Z
sin x
du
tan x dx =
dx =
cos x
u
= ln |u| + C = ln | cos x| + C . sin x dx The Substitution Rule for Definite Integrals Theorem (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 0 f (g (x))g (x) dx =
a Z g (b) f (u) du.
g (a) Integration by parts
Recall the method of substitution 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
0
f (x)g (x) dx = 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 f 0 (x)g (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.
Z
Z
So the formula becomes:
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. R
R
Examples using strategy: u dv = uv
v du
Z
xe x dx : Choose u = x and dv = ex dx
Z
Z
Z
Z t 2 e t dt : Choose u = t2 and dv = et dt
ln x dx : Choose u = ln x and dv = dx
x sin x dx : u = x and dv = sin x dx
x 2 sin 2x dx : u = x2 and dv = sin 2x dx R u dv = uv
Example
Find R v du Z xe x dx. R u dv = uv
Example
Find R v du Z xe x dx. Solution:
Let dv = ex dx. u=x Then
du = dx
Integrating by parts gives
Z
xe x dx = xe x Z v = ex . e x dx = xe x ex + C . Example
Evaluate Z ln x dx. Example
Evaluate Z ln x dx. Solution:
Let Then u = ln x du = dv = dx. 1
dx
x Integrating by parts, we get
Z
ln x dx = x ln x
= x ln x Z v = x. Z dx = x ln x x dx
x x + C. Definition
Let n be a positive integer. Then the Cartesian product of n copies of
the real number line R is:
Rn = R ⇥ R ⇥ . . . ⇥ R = {(a1 , a2 , . . . , an ) | aj 2 R},
which is the set of all ordered n-tuples of real numbers.
Example
(a) R2 = R ⇥ R = {(a1 , a2 ) | ai 2 R} is the Euclidean plane. (b) R3 = R ⇥ R ⇥ R = {(a1 , a2 , a3 ) | ai 2 R} is Euclidian three-space. Notation and coordinates
There are two standard notations for points in R3 , or more generally
Rn . If P 2 R3 , then P = (a1 , a2 , a3 ) for some scalars a1 , a2 , a3 2 R.
The book also denotes this point by writing P(a1 , a2 , a3 ).
The scalar a1 is called the x-coordinate of P, a2 is called the
y -coordinate of P and a3 is called the z-coordinate of P.
Example
The point P = (1, 0, 7) in R3 can also be written as P(1, 0, 7). Its
z-coordinate is 7. The coordinate axes and finding coordinate points The coordinate axes and finding coordinate points Plotting the points ( 4, 3, 5) and (3, 2, 6) Example
What surfaces in R3 are represented by the following equations?
(a) z = 3
(b) y = 5 Example
What surfaces in R3 are represented by the following equations?
(a) z = 3
(b) y = 5
Solution
(a) z = 3 is the set {(x, y , z) | z = 3}, which is the set of points in R3
whose z-coordinate is 3. This is the horizontal plane parallel to the
xy -plane and 3 units above it as in Figure (a).
(b) y = 5 is the set of all points in R3 whose y -coordinate is 5. This is
the vertical plane parallel to the xz-plane and 5 units to the right of
it as in Figure (b). Example
Describe and sketch the surface in R3 represented by the equation y = x. Example
Describe and sketch the surface in R3 represented by the equation y = x.
Solution
The equation represents the set of points in R3 whose x- and
y -coordinates are equal: {(x, x, z) | x 2 R, z 2 R}. This is a vertical plane that intersects the xy -plane in the line
y = x, z = 0.
The portion of this plane that lies in the first octant is sketched
below. Distance Formula in Three Dimensions
The d...

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