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Title: Advanced Calculus - Integration
Description: 15 pages of notes written using a professional PDF writer, producing high quality. There is theory, proofs and examples giving a good understanding of the topic for the reader.
Description: 15 pages of notes written using a professional PDF writer, producing high quality. There is theory, proofs and examples giving a good understanding of the topic for the reader.
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Advanced Calculus - Integration
George Beresford
November 28, 2013
1
Contents
1 Integration
1
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1
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1 Example 1
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1 Proof of property 3
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2
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1 Example 3
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3
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3 Example 4
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4 The area function
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5 The fundemental theorem of Calculus
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2 Example 6
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6 Proof of the fundemental theorem of Calculus
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1
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2 Part 2
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7 Some elementary antiderivatives
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8 Integration by substitution
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2 Example 7
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9 Trigonometric identities
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3 Example 9
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5 Example 11
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1 Example 13
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1 Example 15
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3 Example 17
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12 Improper Integrals
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2 Example 19
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1
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1
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1 Example - Riemann Integrals
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15
1
1
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A partition of the interval [a, b] is
a set of real numbers
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, xn − 1, xn = b} with x0 < x1 <
...
Given a
partition P , we define the left Riemann sum to be:
n−1
n−1
f (xi )(xi+1 − xi ) =
SL =
i=0
f (xi )δ(xi )
i=0
and the right Riemann sum to be:
n−1
n−1
f (xi+1 )(xi+1 − xi ) =
SR =
i=0
f (xi+1 )δ(xi )
i=0
We define the mesh of a partition to be the maximum of the lengths:
δx0 , δx1 ,
...
We must allow the partition to vary, meaning that the number of rectangles,
n, varies also
...
b
f (x) dx
a
1
...
1
Example 1
Consider the interval [0, 10] and a partition P = {0, 1,
...
Write P = {xi : xi = i, i =
0,
...
The left Riemann sum corresponding to the above partition
is:
9
9
f (xi )(xi+1 − xi ) =
SL =
i=0
n
[N
...
3
1
...
2
Example 2
Consider the function f (x) = 1 if 0 ≤ x ≤ 1, and f (x) = x + 1 if 1 < x ≤ 2, and the partition of n + 1
equally spaced points: x0 = 0, x1 = 2(1)/n, x2 = 2(2)/n,
...
It makes it easier if we split the partition into two parts
seeing as there are two parts to our piecewise function
...
, xm−1 , xm
where 0 ≤ xi ≤ 1 and 0 ≤ i ≤ m
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, xn−1 , xn
where 1 < xi ≤ 2 and m + 1 < i ≤ n
...
Supposing n is an even number, n = 2k,
then m = 2k/k = 2, we get this from looking at the maximum value in that section of the function
...
, xk−1 , xk
xk+1 , xk+2 ,
...
4
i)
i=1
1
...
Thet set of inegrable functions fom [a, b] → R
is a vector space, the below rules show that integration is a linear transformation in this vector space
...
B
...
2
...
Suppose P1 : a = x0 < x1 <
...
<
ym−1 < ym = b is a partition of [c, b]
...
< xn−1 < y0 < y1 <
...
The mesh(P ) = max(mesh(P1 ), mesh(P2 ))
...
< n + m where zi = xi if 0 ≤ i ≤ n − 1 and zi = yi−n if n ≤ i ≤ n + m
...
Let the mesh
of the partition mesh(P ) tend to 0, therefore mesh(P1 ) → 0 and mesh(P2 ) → 0
...
2
...
3
f (x) dx
a
The Mean Value Theorem
Let f : [a, b] → R be a conitinuous function, then there exists c ∈ (a, b) so (a < c < b) such that:
1
b−a
b
f (x) dx = f (c)
a
Thinking of integration as a summation process, the mean value theorem states that f attains its
mean value at some point c ∈ (a, b)
...
3
...
The mean value
theorem states that there exists a c ∈ (−90, 90) such that:
1
90 − (−90)
90
f (θ) dθ =
−90
1
180
90
f (θ) dθ = f (c)
−90
The left hand side is the mean temperature on the Greenwich Meridean
...
6
1
...
2
Sketch Proof
Let f : [a, b] → R be a continuous function, m be the minimum value that f attains on [a, b] and M be
the maximum value that f attains on [a, b]
...
The rectangle of
height m from x = a to x = b lies below the curve f , seeing as m is the minimum point in the range
...
Similarly, the rectangle of height M lies above the curve f , seeing as M is
the maximum point in the range
...
Combining the equations results in the
following result
...
3
...
a
Example 4
One may compute that
we have:
10
(x + 1) dx
0
= 60
...
The mean value theorem tells us that there exists some c ∈ (0, 10) such that 6 = c + 1, meaning
that c = 5 in this case
...
4
The area function
Let f : [a, b] → be an integrable function
...
B
...
If it isn’t the area
function cannot be computed]
1
...
(2) If F is a function such that F (x) = f (x) then,
b
f (x) dx = F (b) − F (a)
a
7
1
...
1
Example 5
x
x
(t
a
1
We can show that a (t + 1) dt = 2 (x2 − a2 ) + (x − a)
...
Differentiating A(x) we get,
2
+ 1) dt =
1
A (x) = (2)(x) + 1 = x + 1 = f (x)
2
1
...
2
Example 6
Let f (x) = 3x2
...
Part (2) of the fundemental
theorem of Calculus states,
b
3x2 dx = F (b) − F (a) = b3 − a3
a
1
...
F is also known as the primitive integral or the indefinite integral
...
6
...
A(x + h) − A(h)
= lim f (c) = lim f (ch ),
h→0
h→0
h→0
h
A (x) = lim
With x < ch < x + h
...
1
...
2
Part 2
b
x
If F (x) = f (x) then, a f (x) dx = F (b) − F (a)
...
Introducing the equation G(x) = F (x)−A(x) =⇒ G (x) = F (x)−A (x) =⇒ G(x) = f (x)−f (x) = 0
...
So F (x) = A(x) + G(x) =⇒ F (x) = A(x) + c
...
To conclude, the fundemental theorem of Caculus says (1) differentiation is the inverse of integration
...
8
1
...
cos(x) dx = −sin(x) + c
sin(x) dx = −cos(x) + c
sec2 (x) dx = tan(x) + c
cosec2 (x) = −cot(x) + c
xn dx =
√
xn+1
+c
n+1
1
dx = sin−1 (x) + c
1 − x2
1
dx = tan−1 (x) + c
1 + x2
ex dx = ex + c
1
dx = log(x) + c
x
ax dx =
1
ax + c
log(a)
cosh(x) dx = sinh(x) + c
sinh(x) dx = cosh(x) + c
1
...
8
...
Define u(x) = 1 + x2 and u (x) = 2x
...
√
Then f (u(x)) = u(x) = 1 + x2
...
B
...
8
...
e
...
Recall the
x
area function A(x) = 0 g(t) dt is an antiderivative of g(x)
...
1
1
g(t) = t3 cos(2 + t4 ) = (4t3 )cos(2 + t4 ) = u (t)f (u(t))
4
4
where f (t) = cos(t)
...
1
...
9
...
Consider ei(x+y) = cos(x + y) + isin(x + y)
...
eiy
eix
...
9
...
0
Now, sin2 (x) = 1 (1 − cos2x)
...
9
...
Now sin4 (x) = (sin2 x)2 = ( 1 (1 − cos2x))2
...
9
...
9
...
First, look at the formula sin2 θ + cos2 θ = 1
...
Instead take x = 3cosθ so x2 = 9cosθ
...
So,
√
9 − x2
3sinθ
I=
dx =
...
I = cos−1 (x/3) − tan(cos−1 (x/3)) + c
Using tan2 y = sec2 y − 1 to simplify tan(cos−1 (x/3))
...
9
...
A substitution of x = acostheta will not work here
√
√
as we have x2 − a2 instead of a2 − x2
...
We make the substitution
x = acoshθ then,
x2 − a2 = a2 cosh2 θ − a2 = a2 (cosh2 θ − 1) = a2 sinh2 θ
x (θ) = asinhθ =⇒ I =
1
...
10
...
2x2 −x+4
x(x2 +4)
dx
...
Therefore,
x2
1
dx =
+4
1
1
dx =
+4
4
4u2
u2
1
1
dx =
+1
2
u2
1
1
1
du = tan−1 u + c = tan−1 (x/2) + c
+1
2
2
1
Through a similar process x2x = 2 log(x2 + 4) + c, however using the substitution u = x2 + 4 instead
...
2
In the above example the degree of the numerator is less than the degree of the denominator
...
1
...
2
Example 14
Compute the following indefinite integral
...
I =x+
x−1
dx = x +
(2x − 1)2 + 2
1
(u
2
+ 1) − 1
1
du = x + (
2 + 2)
2(u
4
u2
u
1
− 2
)
+2 u +2
1
The first half of the integral integrates to 2 log(u2 + 2) using the substitution v = u2 + 2
...
Theresfore I is:
√
√
1 1
1 2
1
2
2x − 1
2
−1 u
2
I = x + ( log(u + 2)) − (
tan ( √ )) + c = log((2x − 1) + 2) −
tan−1 ( √ ) + c
4 2
4 2
8
8
2
2
12
1
...
Integrating all the terms results
b
b
d
u(x)v(x) dx =
dx
u(x)v (x) dx +
a
Write F (x) = u(x)v(x) then F (x) = f (x) where f (x) =
d
(u(x)v(x))
...
11
...
logt dt
1
1
u(t) = logt, u (t) = , v (t) = 1, v(t) = t
t
x
x
x
1
1
1
1 dt = (xlogx − 1log1) − [t]x = xlogx − (x − 1)
1
u v dt = [tlogt]x −
1
uv dt = [uv]x −
1
I=
x(logx − 1) + 1
1
...
2
Example 16
Integrate the indefinite integral
I=
x2 ex dx
u(x) = x2 , u (x) = 2x, v (x) = ex , v(x) = ex
I = [x2 ex ] −
2xex dx = x2 ex − 2
xex dx
Integrating by parts again
u(x) = x, u (x) = 1, v (x) = ex , v(x) = ex
[xex ] −
ex dx = xex − ex + c =⇒ I = ex (x2 − 2x + 2) + c
With indefinite integrals we can check using the fundemental theorem of Calculus to see whether the
integral I is correct
...
11
...
1
...
We can say,
b
f (x) dx
a
where [a, b] is finite and f i continuous
...
12
...
12
...
B
...
12
...
5
√
I=
2
1
dx = lim
t→2+
x−2
5
√
t
1
dx
x−2
5
(x − 2)−1/2 dx = [2(x − 2)1/2 ]5 = 2(3)1/2 − 2(t − 2)1/2
t
t
√
√
I = lim [2(3)1/2 − 2(t − 2)1/2 ] = [2 3 − 2(2+ − 2)1/2 ] = 2 3
+
t→2
1
...
13
...
We’ll find a Riemann sum approximation
7
to 4 f (x) dx
...
n−1
f (xi )(xi+1 − xi ) = f (x0 )(x1 − x0 ) + f (x1 )(x2 − x1 ) = (2
...
5 + 3) = 37
SL =
i=0
Here SL underestimates the integral
...
3
3
Each interval will have a width 7−4 = n
...
, xn }, then we have x0 = 4, x1 = 4 + n ,
n
x2 = 4 + 3
...
i , xn = 4 + 3
...
The left Riemann sum is therefore,
n
n
n
n−1
SL =
i=0
n−1
i=0
n−1
3
f (xi )(xi+1 − xi ) =
(2xi + 3)( ) =
n
i=0
33
33 18
+ 2 =
n
n
n
n−1
i=0
18
1+ 2
n
n−1
i=
i=0
n−1
3
...
n n
33
18 1
9
9
n + 2 ( (n − 1)(n)) = 33 + (n − 1) = 42 −
n
n 2
n
n
As n → ∞, SL → 42, so its explicit that SL underestimates the Riemann integral
...
1
...
2
Example - Integration by substitution
π/4
Integrate 0 esinx cosx dx
...
So if f (x) = ex , we have f (u(x)) = es inx
Title: Advanced Calculus - Integration
Description: 15 pages of notes written using a professional PDF writer, producing high quality. There is theory, proofs and examples giving a good understanding of the topic for the reader.
Description: 15 pages of notes written using a professional PDF writer, producing high quality. There is theory, proofs and examples giving a good understanding of the topic for the reader.