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MathematicsLecture and practice
Norbert Bogya
University of Szeged, Bolyai Institute
2017
Norbert Bogya (Bolyai Institute) Mathematics 2017 1 / 23
Integration
Norbert Bogya (Bolyai Institute) Mathematics 2017 2 / 23
Motivating question
What if?
We know the velocity function of an object, but we want toknow the position function.
We want to recover a function from its known rate of change.
Definition
A function F is an antiderivative of f on an interval I ifF ′(x) = f (x) for all x ∈ I .
The process of recovering a function F (x) from its derivative f (x) iscalled antidifferentiation.
Norbert Bogya (Bolyai Institute) Mathematics 2017 3 / 23
Antiderivatives
Exercise
Find an antiderivative for each of the following functions.
(a) f (x) = 2x
(b) g(x) = 5x
(c) h(t) = 4t2 − 3t7
(d) h(x) =√x3 + 2
5√x2
Theorem
If F is an antiderivative of f on an interval I , then the most generalantiderivative of f on I is
F (x) + C
where C is an arbitrary constant.
Norbert Bogya (Bolyai Institute) Mathematics 2017 4 / 23
Antiderivatives
Function Antiderivativec c · x
xα (α 6= −1)xα+1
α + 11/x log |x | (ln |x |)sin x − cos xcos x sin xex ex
Norbert Bogya (Bolyai Institute) Mathematics 2017 5 / 23
Particular antiderivative
The previous theorem provides that the most general antiderivative off on I is a family of functions whose graphs are vertical translates ofone another. We can select a particular antiderivative from thisfamily by assigning a specific value to C .
Exercise
Find an antiderivative of f (x) =x3 − 2x
xthat satisfies F (1) = 2.
Norbert Bogya (Bolyai Institute) Mathematics 2017 6 / 23
Linear substitution
Theorem
If F (x) is an antiderivative of f (x), thenF (ax + b)
ais an
antiderivative of f (ax + b).
Example
f (x) =√
3x − 5, F (x) =23(3x − 5)3/2
3
Exercise
Find the general antiderivative of each of the following functions.
(a) f (x) =√
2x − 1(b) g(x) = 5
√4x − 3
(c) h(x) =6
(5− x)2
Norbert Bogya (Bolyai Institute) Mathematics 2017 7 / 23
Indefinite integral
Definition
The set of all antiderivatives of f is the indefinite integral of f withrespect to x , denoted by ∫
f (x)dx .
The symbol∫
is an integral sign. The function f is the integrand ofthe integral, and x is the variable of integration.
Example
(a)∫
(x + 1)dx =x2
2+ x + C , C ∈ R
(b)∫ √
2x − 1dx =23(2x − 1)3/2
2+ C , C ∈ R
Norbert Bogya (Bolyai Institute) Mathematics 2017 8 / 23
Definite integral - Motivating question
What is the area of the shadedregion R that lies above thex-axis, below the graph ofy = 1− x2 and between thevertical lines x = 0 and x = 1?
Norbert Bogya (Bolyai Institute) Mathematics 2017 9 / 23
Approximating Area
The area of a region with a curved boundary can be approximated bysumming the areas of a collection of rectangles. Using morerectangles can increase the accuracy of the approximation.
A(a) ≈ 1 · 12
+ 34· 12
= 78
= 0.875 A(b) ≈ 0.78125Norbert Bogya (Bolyai Institute) Mathematics 2017 10 / 23
Approximating Area
Lower estimation
A ≈ 0.634765625
Upper estimation
A ≈ 0.697265625
The exact area is between the lower and the upper estimation.
Norbert Bogya (Bolyai Institute) Mathematics 2017 11 / 23
Notation
“Integral of f from a to b”∫ b
af (x) dx
a: Lower limit of integrationb: Upper limit of integrationf (x): the function is the integranddx : x is the variable of integration
Norbert Bogya (Bolyai Institute) Mathematics 2017 12 / 23
Definite integral
Definition
Geometrically∫ b
af (x)dx is the signed area of the region that lies
between the x-axis, the graph of y = f (x) and the vertical lines x = aand x = b. Evaluation of the integral means finding the exact area.
Natural question: can this definite integral evaluated for everyfunction f (x)? (No.)
Theorem
A continuous function is integrable. That is, if a function f iscontinuous on an interval [a, b], then its definite integral over [a, b]exists.
Norbert Bogya (Bolyai Institute) Mathematics 2017 13 / 23
Exercise
x
y
11
f
Find
∫ 2
−2f (x)dx .
x
y
11
f
A1 = 4 and A2 = −1.5∫ 2
−2f (x)dx = 4− 1.5 = 2.5
Norbert Bogya (Bolyai Institute) Mathematics 2017 14 / 23
Properties of definite integrals
Norbert Bogya (Bolyai Institute) Mathematics 2017 15 / 23
Mean value theorem for definite integrals
Theorem
If f is continuous on [a, b], thenat some point c ∈ [a, b],
f (c) =1
b − a
∫ b
a
f (x)dx .
Definition
The number1
b − a
∫ b
a
f (x)dx is called the integral mean (or
average) of f on the interval [a, b].
Norbert Bogya (Bolyai Institute) Mathematics 2017 16 / 23
Exercise
x
y
11
f
Find the integral mean over the interval [−2, 2].
f [−2,2] =
∫ 2
−2 f (x)dx
2− (−2)=
2.5
4= 0.625
Norbert Bogya (Bolyai Institute) Mathematics 2017 17 / 23
Connection between definite integral and
differentiation (Fundamental theorem of calculus)
F (x) =∫ x
af (t)dt
a ≤ x ≤ b
Theorem
If f is continuous on [a, b] thenF (x) =
∫ x
af (t)dt is continuous
on [a, b] and differentiable on(a, b) and its derivative is f (x);
F ′(x) = f (x).
Norbert Bogya (Bolyai Institute) Mathematics 2017 18 / 23
Connection between definite integral and
differentiation (Fundamental theorem of calculus)
Theorem
If f is continuous at every point of [a, b] and F is any antiderivativeof f on [a, b], then ∫ b
a
f (x)dx = F (b)− F (a).
This theorem is also known as formula of Newton-Leibniz.
Norbert Bogya (Bolyai Institute) Mathematics 2017 19 / 23
Examples
∫ 2
−3(6− x − x2)dx =
[6x − x2
2− x3
3
]2−3
= (12− 2− 8
3)− (−18− 9
2+
27
3) =
125
6
∫ 4
1
(3
2
√x − 4
x2
)dx =
[x3/2 +
4
x
]41
=
(43/2 +
4
4
)︸ ︷︷ ︸
8+1=9
−(
13/2 +4
1
)︸ ︷︷ ︸
1+4=5
= 4
Norbert Bogya (Bolyai Institute) Mathematics 2017 20 / 23
Total area
IMPORTANT
Definite integral is not the total area.
To find the area between the graph of y = f (x) and the x-axis overthe interval [a, b], do the following.
(1) Subdivide [a, b] at the zeros of f .
(2) Integrate f over each subinterval.
(3) Add the absolute values of the integrals.
Exercise
Find the total area of the region between the x-axis and the graph off (x) = x3 − x2 − 2x , −1 ≤ x ≤ 2.
Norbert Bogya (Bolyai Institute) Mathematics 2017 21 / 23
Physics
Exercise
The velocity of a moving body is described by the function
v(t) = t2 − t(ms
).
(a) Find the indefinite integral of v(t).
(b) Determine the position function s(t), such that s(0) = 1.
(c) Calculate the average velocity over the interval [1, 4].
(d) Find the extremal value of s(t).
(e) Find the acceleration function of the body.
Norbert Bogya (Bolyai Institute) Mathematics 2017 22 / 23
Physics
Solutions:
(a)∫v(t)dt = t3
3− t2
2+ C
(b) s(t) = t3
3− t2
2+ 1
(c) v [1,4] =∫ 41 v(t)dt
4−1 = s(4)−s(1)3
=
(43
3− 42
2
)−( 1
3− 1
2)3
(d) Position function s(t) has extremum if and only if v(t) = 0 andv(t) changes sign.
v(t) = 0⇐⇒ t = 1 or t = 0
t < 0 t = 0 0 < t < 1 t = 1 1 < t
v + 0 − 0 +
s ↗ maxs(0)=1 ↘ min
s(1)=5/6 ↗(e) a(t) = 2t − 1
Norbert Bogya (Bolyai Institute) Mathematics 2017 23 / 23
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