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      Calculus II

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© The scientific sentence. 2010

Calculus II:
Definite integral
Sum and integral



1. Introduction

1.1. Evaluating a distance



An object moves at a speed v(t) = at + vo (m/s) from a point a at the time ta to a point b at the time tb in an interval [a,b].

What is the value of the traveled distance Xab ?

If the speed were constant, the problem would be simple and we could write: Xab = v (tb - ta). But it is not.

We start to subdivide the interval [a,b] in some sub-intevals, let 5 sub-intervals. We have, therefore, 6 points a=x0, x1, x2, x3, x4, b=x5.

We consider the midpoint mi of each sub-interval, we have, therefore, 5 points m0, m1, m2, m3, m4, with
mi = ti + (ti+1 - ti)/2 = (ti+1 + ti)/2 = ti + Δti/2.

Inside each sub-interval, we consider the speed constant equal to v(mi) the speed at the midpoint mi. So for each sub-interval [xi,xi+1], we write:

Δxi = xi+1 - xi = vi . Δti,

Where vi = v(mi) and Δti = ti+1 - ti

Therefore

The distance Xab = Σ vi . Δti
i = 0, 1, 2, 3, 4, 5.

Xab = v0(t1 - t0) + v1(t2 - t1) + v2(t3 - t2) + v3(t4 - t3) + v4(t5 - t4) =

v(m0)(t1 - t0) + v(m1)(t2 - t1) + v(m2)(t3 - t2) + v(m3)(t4 - t3) + v(m4)(t5 - t4)

We substitute in the above formula the expressions of vi = v(mi) = a mi + vo we get:

Xab = (a m0 + vo)(t1 - t0) + (a m1 + vo)(t2 - t1) + (a m2 + vo)(t3 - t2) + (a m3 + vo)(t4 - t3) + (a m4 + vo)(t5 - t4) =

a[m0(t1 - t0) + m1(t2 - t1)+ m2(t2 - t3) + m3(t4 - t3) + m4(t5 - t4)] + vo[(t1 - t0) + (t2 - t1) + (t3 - t2) + (t4 - t3) + (t5 - t4)] =

a[m0(t1 - t0) + m1(t2 - t1)+ m2(t2 - t3) + m3(t4 - t3) + m4(t5 - t4)] + vo[t5 - t0]

With mi = (ti+1 + ti)/2, we obtain:

Xab = (a/2)[t12 - t02 + t22 - t12 + t32 -t22 + t42 - t32 + t52 - t42] + vo[t5 - t0] =

(a/2)[t52 - t02] + vo[t5 - t0] =
[(a/2)(t5 + t0) + vo][t5 - t0]

If all the periods of time within the sub-intervals are equal to T, that is Δti = ti+1 - ti = T
We will have:

t5 = t4 + T = t3 + 2T = t2 + 3T = t1 + 4T = t0 + 5T

so

t5 + t0 = 2t0 + 5T
t5 - t0 = 5T

so

Xab = 5T[(a/2)(2t0 + 5T) + vo]

With n subdivisions, we have

n = (b - a/Δxi) = (x5 - x0/Δxi) =
(t5 - x0)/Δti) = (tb - ta)/T


The gneral formula is:


Xab = nT[(a/2)(2t0 + nT) + vo]



Example:
ta = 0
tb = 10
T = 1 s,
vo = 3 m/s,
a = 2 m/s/s,

Then
n = (10 - 0)/1 = 10
Xab = 10[(2/2)(0 + 10) + 3] = 130 m.



1.2. Evaluating an area



Now we want to evaluate the area delimited by the curve of equation y = f(x) = x2 + 1, the x-axis, the y-axis, and the vertical line x = 5.

The related area doesn't have a simple and regular geometric form. We will evaluate this area by approximation and divide this area in five small rectangles.

We sub-divide the interval [0,5] in 5 sub-intervals of width equal to 1 each. We consider then the midpoint of each sub-interval

We construct then five rectangles of base equal to the width of each sub-interval, that is 1 = Δxi = xi+1 - xi , and of height the image of the midpoint by the function f(mi), woth mi = (xi+1 + xi)/2

We calculate the area of each rectangle and sum the areas of the five rectangles.

Ai = [xi+1 - xi][f(mi)] with mi = (xi+1 + xi)/2

For the first rectangle:

A1 = [x1 - x0][f(m0)]
with
[x1 - x0] = 1
m0 = (x1 + x0)/2 = (1 + 0)/2 = 1/2
so A1 = 1 . f(1/2) = 1. ((1/2)2 + 1) = 5/4

For the second rectangle:

A2 = [x2 - x1][f(m1)]
with
[x2 - x1] = 1
m1 = (x2 + x1)/2 = (2 + 1)/2 = 3/2
so A2 = 1 . f(3/2) = 1. ((3/2)2 + 1) = 13/4

Similarly

A3 = 1 . f(5/2) = 1. ((5/2)2 + 1) = 29/4

A4 = 1 . f(7/2) = 1. ((7/2)2 + 1) = 53/4

A5 = 1 . f(9/2) = 1. ((9/2)2 + 1) = 85/4

The area of the rectangle is then

A = A1 + A2+ A3 + A4 + A5
= 5/4 + 13/4 + 29/4 + 53/4 + 85/4 =
(5 + 13 + 29 + 53 + 85)/4 = 185/4 = 46.25

The general formula is:


A = Σ Ai = Σ f(mi) Δxi
mi = (xi+1 + xi)/2
Δxi = xi+1 - xi





1.3. Evaluating a volume



We vave the following formulas:

V1 = πr12 h1
V2 = πr22 h2
V3 = πr32 h3
...
V10 = πr102 h10

V = Σ Vi = V1 + V2 + V3 + ... + V10
= πr12 h1 + πr22 h2 + πr32 h3 + ... + πr102 h10

The subdivision has the same lenghth h, so

V = π h [r12 + r22 + r32 + ... + r102] =
π h Σri2
i = 1 → 10

r2 = r1 - a
r3 - r2 - a = r1 - 2a
r4 = r1 - 3a
...
r10 = r1 - 9a


Therefore

Σri2 = r12 + r22 + r32 + ... + r102 =
r12 + (r1 - a)2 + (r1 - 2a)2 + ... + (r1 - 9a)2 =
10 r12 - 2ar1(1 + 2 + 3 + ... + 9) + a2(12 + 22 + 32 + ... + 92)

Using the formulas:

Σ i = n(n + 1)/2
i = 1 → n

Σ i2 = n(n + 1)(2n + 1)/6
i = 1 → n


We have

1 + 2 + 3 + ... + 9 = 9 x 10/2 = 45

12 + 22 + 32 + ... + 92 = 9 x 10 x 19/6 = 285


Therefore

Σri2 = 10 r12 - 2ar1 x 45 + 285 a2
= 10 r12 - 90 a r1 + 285 a2

Hence

V = π h (10 r12 - 90 a r1 + 285 a2)

Since r1 = 11 a

We obtain:

V = π h (10 r12 - (90/11) r12 + (285/121)r12) =
π h r12(10 - (90/11) + (285/121)) =
4.17 π h r12

Since H = 10 h, we get:

V = 0.417 π H r12

With n subdivisions, the general formula is:

V = π h [r12 + r22 + r32 + ... + rn2]
= π h [ n r12 - 2ar1(n-1)n/2 + a2(n-1)(n)(2n - 1)/6]

r1 = (n+1)a

Therefore

V = π h [ n r12 - 2r12(n-1)n/2(n + 1) + r12(n-1)(n)(2n - 1)/6(n + 1)2]

With nh = H and r1 = r, we have:

V = π H r2[1 - (n-1)/(n + 1) + (n-1)(2n - 1)/6(n + 1)2]


V = π H r2[1 - (n-1)/(n + 1) + (n-1)(2n - 1)/6(n + 1)2]




With an infinite subdivisions:

lim V = lim π H r2[1 - (n-1)/(n + 1) + (n-1)(2n - 1)/6(n + 1)2]
n → ∞

= π H r2 lim [1 - (n-1)/(n + 1) + (n-1)(2n - 1)/6(n + 1)2] =
n → ∞

= π H r2[1 - 1 + 2/6] = π H r2/3


V (cone) = π H r2/3



That is the volume of the cone of height H and radius r.



4. Exercises






  


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