Download CHAPTER C2

Ms. Amal J. El-sayed
MST(121)
CHAPTER C2.
Integration and Modeling
The process of undoing differentiation is called integration. The outcome of
integration is called an integral.
Now ∫f(x). dx = f(x) + c
where c is constant, ∫f(x) . dx " is called
indefinite integral".
A table of integrals:
1. f (x) = a (a is constant) → ∫f(x) dx = ∫a. dx = ax + c
2. f (x) = xn (n is constant) → ∫f(x) dx = ∫xn. dx = (1∕n+1 . xn+1) + c
3. f (x) = 1/x (x>0) → ∫1/x. dx = ln x + c
4. f (x) = eax → ∫eax dx = (1/a) eax + c
5. f (x) = Cos(ax) → ∫Cos(ax) . dx = (1/a) Sin(ax) + c
6. f (x) = Sin(ax) → ∫Sin(ax) . dx = –(1/a) Cos(ax) + c
7. ∫lnx. dx = x(lnx–1) + c
Example: Find the indefinite integral of the function f(t) = 2t
Solution: ∫f(t)dt = ∫2t.dt = 2∫t.dt = 2(t2/2) + c = t2 + c .
Examples:
1. What is ∫Cos(x). dx
Solution: ∫Cos(x). dx = Sin(x)+c
2. What is ∫1.dx?
Solution: ∫1.dx = x+c
3. What is ∫3x2.dx?
Solution: ∫3x2.dx = 3(x3/3) +c = x3+c
4. What is ∫n xn-1. dx ?
Solution: ∫n xn-1. dx = n(1/n . xn ) + c = xn+c
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5. What is ∫1/x. dx ?
Solution: ∫1/x . dx = lnx+c
6. What is ∫x4 . dx ?
Solution: ∫ x4.dx = (1/5. x5 ) + c
7. What is ∫Sin(x). dx ?
Solution: ∫ Sin(x). dx = –Cos(x)+c
8. What is ∫e3x. dx ?
Solution: ∫e3x. dx = (1/3. e3x) + c
9. What is ∫(e3x + Sin6x). dx ?
Solution: ∫(e3x + Sin6x). dx = (1/3. e3x) – 1/6 Cos(6x) = c
10.What is ∫5Cos(x). dx ?
Solution: ∫5Cos(x). dx = 5∫Cos(x). dx = 5Sin(x) + c
11.What is ∫(x2 + 3x). dx ?
Solution: ∫(x2 + 3x). dx = 1/3x3 + 3/2x2 + c
12.What is ∫ x . dx ?
Solution: ∫ x . dx = x1/2. dx = [1/(1/2+1)] x1/2+1 + c = 2/3 x3/2+c
13.What is ∫(9x-1/2 + 5x3/4 – 2x+1/x). dx ?
Solution: ∫(9x-1/2 + 5x3/4 – 2x+1/x). dx
= 9/(1/2). x1/2 + 5/(7/4). x7/4 – 2/2x2 + lnx + c
= 18 x1/2 + (20/7) x7/4 – x2 + lnx + c
14.What is ∫x(x3 + 2x-2/3 – 1/x). dx ?
Solution: ∫x(x3 + 2x-2/3 – 1/x). dx
= ∫(x4 + 2x1/3 – 1). dx
= 1/5x5 + [2/(4/3)]x4/3 – x + c
= 1/5x5 + 6/4x4/3 – x + c
! Do activity 1.4
2
The functions Sinh and Cosh are defined by:
Cosh x = (ex+e-x)/2
and Sinh x = (ex–e-x)/2
Find the indefinite integral ∫Cosh x . dx
Answer: ∫Cosh x . dx = ∫(ex+e-x)/2 . dx = 1/2∫(ex+e-x).dx
= 1/2(ex–e-x) + c = Sinh x + c
Activity 1.6 : Find the indefinite integrals:
∫Cos2x . dx and ∫Sin2x . dx
Solution : ∫Cos2x . dx
Note that Cos2x = Cos2x – Sin2x and 1 = Cos2x + Sin2x
Add the 2 equations, So Cos2x + 1 = 2Cos2x
Hence 1/2(Cos2x + 1) = Cos2x
So ∫Cos2x . dx = 1/2∫(Cos2x + 1)dx = 1/2 [1/2Sin2x + x] + c
Therefore ∫Cos2x . dx = 1/4Sin2x + 1/2x + c
Now Do ∫Sin2x . dx !
Do Exercise for Section 1.
Section 3 Differential equations and applications.
The equation of the form dy/dx = f(x) is called the differential
equation. [d(dependent variable) / d(independent variable)] = f(x)
(x- Independent variable)
The Solution of this equation is the integration of the given function.
Activity 3.1
Find the general solution of each of the following differential
equations:
a) dp/dt = Cost + 2
b) dy/dx = e4x + e-4x
c) dz/du = 3Sin(wu)
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Answer :
a) p = ∫(Cost + 2)dt = Sint + 2t + c
b) y = ∫(e4x + e-4x)dx = 1/4 e4x – 1/4e4x + c
c) ∫3Sin(wu)du = –3/w . Cos(wu) + c
Activity 3.3 Solving initial value problems. Using your answers to
Activity 3.1, Solve each of the following initial value problems.
a) dp/dt = Cost + 2
p = 4 when t = 0
b) dy/dx = e4x + e-4x
y = –1 when x = 0
c) z(m) = 3Sin(wu)
z (0) = w-1 (where w is constant)
Solution:
a) from activity 3.1 we got p = Sint + 2t + c
To find c solve 4 = Sin(0) + 2(0) + c
4=0+0+c→c=4
So p = Sin t + 2t + 4
b) From activity 3.1 y = 1/4 e4x – 1/4e-4x + c
Now
–1 = 1/4 e4(0) – 1/4e-4(0) + c
–1 = 1/4 – 1/4 + c → –1 = c
Hence y = 1/4 e4x – 1/4e-4x – 1
c) From activity 3.1 z = –3/w . Cos(wu) + c
w-1 = –3/w . Cos(w(0)) + c
w-1 = –3/w + c → 1/w = –3/w + c
→ c = 1/w + 3/w = 4/w
Therefore
z = –3/w Cos(wu) + 4/w
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3.2 Motion with Constant acceleration.
Given dv/dt = a (a is constant acceleration)
So v = ∫a.dt = at + c
Where c is the value of v when t = 0
1) So v = at + v0
Now ds/dt = v = at + v0
s = ∫v.dt = ∫(at+v0)dt
s = (a/2)t2 + v0t + c where c is the value of s when t = 0
2) Hence s = (a/2)t2 + v0t + s0
When v0 = 0 and s0 = 0 we get
3) s = (a/2)t2 and v = at
4) From first formula we get:
v = at + v0 → (v – v0)/a = t
5) The velocity and position of the object are related by the
equation 2as – v2 = 2as0 – v20
Example 3.1: Dropping a stone.
The height of the Eiffel Tower is 300 meters.
Suppose that a stone is dropped from rest at the top of the tower.
a) Ignoring air resistance, estimate the speed of the stone when it
bits the ground.
b) How long does the stone take to reach the ground?
Solution:
a) Since air resistance is ignored the stone has a constant
acceleration a = g = 9.8 m/s2. The stone is released from rest,
that is with zero velocity at the origin, so we have v0 = s0 = 0
Now 2as0 – v20 = 2as – v2
0 = 2 (9.8) (300) – v2
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→ v = 2.(9.8).300 ≈ 76.7 m/sec.
b) t = (v - v0)/a = (76.7 – 0)/9.8 ≈ 7.8 sec.
so the value of t when the stone hit the ground is 7.8 sec.
Do activity 3.6
3.3 Population growth and radioactive decay:
1) The differential equation dp/dt = kp where k is constant is a
model for population growth.
2) The solution of this equation is p = p0 ekt where p0 is the
initial population and k is the rate of growth.
3) Doubling time: (is the time at which p = 2p0).
Doubling time is t = ln2/k
4) The differential equation dm/dt = –λm where m is the mass of
radioactive substance presents at time (t) is a model for
radioactive decay.
5) The solution of this model is m = m0 e–λt
6) Decay half time t = ln2/λ
Example 3.2 Breeding bacteria.
A colony of bacteria number 1000 at midday on 1 June and its growth
can be modeled by an exponential model with proportionate growth
rate 1 per day.
a) Express the population size as a function of time.
b) What is the size of this population one day later, at midday on
2 June?
Solution:
a) k = 1 and p0 = 1000
So the population size = p = 1000 et
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b) At midday on 2June, the time in days is t =1 and the number of
bacteria has therefore reached p = 1000e (1) = 2720 (to 3 s.f).
Activity 3.8: A human population
The population of a certain country is currently 10 million and is
increasing at 3% per year. Use the exponential model to predict what
the population of the country will be in 10 years time. Find the
doubling time.
Solution:
p0 = 10 million, k = 3% = 0.03
The exponential model is p = p0 ekt
p = 107 e (0.03) t
In 10 years time the population will be
p10 = 107 e(0.03)(10) = 107 e0.3 ≈ 13.5 million
The doubling time t = ln2/0.03 = 23.1 years.
Activity 3.9 Finding particular solutions.
a) Find the particular solution of the differential equation
dy/dx = 2y for which y = 100 when n = 0
b) Find the particular solution of the differential equation
dv/dt = –3v for which v = 15 wfen t = 0
Solution:
a) dy/dx = 2y is a differential equation with k = 2
so the solution is y = y0 ekx
y = y0 e2x
to find y0 use y = 100 when x = 0 so 100 = y0 e2(0) →100=y0(1)
Hence y0 = 100 and so y = 100 e2x.
b) dv/dt = –3v →solution is v= v0e–3t 15= v0e0 →v0=15
so v=15 e–3t
Do activity 3.10 and example 3.4
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Section 4 Definite integrals, areas and summations.
Definition: The definite integral of a continuous function f(x) from a
b
to b defined by

f(x) .dx is defined to be [ F(x) ]ba = F (b) – F (a).
a
Example 4.1

Evaluate the definite integral

Sinx.dx
0

Solution:

Sinx.dx = [– Cosx]0π
= –Cosπ + Cos0 = –(–1) + 1 = 2
0
Do activity 4.4
4
Example: Evaluate the definite integral  1 /u . du
1
4
Solution:
 1 /u . du = [ln u]
= ln4 – ln(1) = ln4.
4
1
1
Finding the area under a graph.
If f(x) is a continuous function which takes no negative values for
a ≤ x ≤ b then the area bounded by the graph of f(x), the x- axis and
b
the lines x=a and x=b is equal to the definite integral
 f (x). dx.
a
Example:
a) What is the area under the graph of f(x) =3 which is bounded
also by the x- axis and by the lines x=0 and x=5.
5
Solution: The area A =  3 .dx = [3x ]50 = 3(5) – 3(0) = 15 unit2.
0
b) What is the area under the graph of f(x) =2x+3 which is
bounded also by the x- axis and by the lines x= –1 and x=2.
8
2
Solution : Area =  2x  3 .dx = [(2/2)x2 + 3x] 21 = [x2 + 3x] 21
1
= [(2)2 + 3(2)] – [(-1)2 + 3(-1)]
= 10 – (-2) = 12 unit2.
Example 4.2:
Find the area under the graph of Sinx from x=0 to x=π

Solution: The area is equal to the definite integral  Sinx .dx= [-Cosx] 0
0
= -Cos(π) – (-Cos(0)) = -(-1) – (-1) = 2 unit2.
Do activity 4.6
Remarks:
1) IF f and g are continuous and f(x) ≥ g(x) for all x in [a,b], then
the area of the region bounded by the graphs of f and g, x = a
b
and x = b is A =   f ( x)  g ( x) .dx.
a
2) Let f be continuous on [a,b]. The average value fav of f on [a,b]
b
is fav = (1/(b-a))
 f ( x ) .dx.
a
Find the area of the region bounded by the parabola y = 1-x2 and
the line y = x-1
Solution: first find the points of intersection of the line and the
parabola.
1-x2 = x-1
0 = x2 + x – 2
0 = (x+2)(x-1)
x = -2 , x = 1
From the graph 1-x2 ≥ x-1 on [-2,1]
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2
So Area =

[(1-x2) – (x-1)].dx = [x – (1/3) x^3 – (1/2)x2 + x] 21
1
= [2x – (1/2x2) – (1/3x3)] 21
= 7/6 + 10/3 = 7/6 + 20/6 = 27/6 = 9/2
Find the area bounded by the function y = x2 + 1 and the line y = 5.
Answer: first find the points of intersection of the line y = 5 and the
parabola.
x2 + 1 = 5 → x2 + 1 – 5 = 0
x2 – 4 = 0 → x2 = 4 → x = ± 4
So x = 2, x = -2
y= 5 ≥ (x2 + 1) on [-2,2]
2
So A =

[5 – (x +1)].dx =
2
2
2

(4 – x2). dx
2
= [4x – (1/3)x2)] 22
= 32/3
Remark: Find the area bounded by the parabola y = 4 – x2 and the x
axis means that.
Find the area bounded by the parabola y = 4 – x2 and the line y = 0.
Definition: Let f be continuous on [a,b]. The average value fav of f on
b
[a,b] is fav = (1/(b-a))
 f ( x ) .dx.
a
Example: Find the average value of f on [a,b]
3
1)

3
3x2.dx , 2)
0

(3x2 – 2x + 3).dx
1
Solution:
3
1) Average value fav = 1/(3-0)  3x2.dx = 1/3 [(3/3)x3)] 30
0
10
= 1/3[x3] 30 = 1/3(27) = 9\
3
2) Average value = 1/(3-(-1))  (3x2 – 2x + 3).dx
1
3
= 1/4  (3x2 – 2x + 3).dx
1
= 1/4 [x3 – x2 + 3x] 31
= 1/4 (27–(–5))
= 1/4 (27+5)
= 32/4
=8
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