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 = t_i + (t_i+1 - t_i)/2 = (t_i+1 + t_i)/2 = t_i + Δt_i/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 [x_i,x_i+1], we write:

Δxi = x_i+1 - x_i = vi . Δti,

Where vi = v(mi) and Δti = t_i+1 - t_i

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 = (t_i+1 + t_i)/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 Δt_i = t_i+1 - t_i = 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:

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 = Δx_i = x_i+1 - x_i , and of height the image of the midpoint by the function f(mi), woth mi = (x_i+1 + x_i)/2

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

Ai = [x_i+1 - x_i][f(mi)] with mi = (x_i+1 + x_i)/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:

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]

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

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

4. Exercises