## Friday, April 28, 2017

### Unizor - Definite Integrals - Area Examples

Notes to a video lecture on http://www.unizor.com

Definite Integrals -
Area Examples

Example 1

Find "area under curve" for f(x) = 10x on segment [a=0, b=4].

Solution

First, let's experiment with a couple of simple cases.

Case 1. N=2
Point of division in two equal parts is x1=2
So, a=x0=0x1=2x2=4=b
Minimum on the first interval [0, 2] is 0 at x=0.
Minimum on the second interval [2, 4] is 20 at x=2.
Therefore, the sum of areas of all rectangles equals to
S1 = 0·2 + 20·2 = 40.

Case 2. N=4
Two additional points of division in four equal parts are x1=1 and x3=3
So, a=x0=0x1=1x2=2,
x3=3x4=4=b
Minimum on the first interval [0, 1] is 0 at x=0.
Minimum on the second interval [1, 2] is 10 at x=1.
Minimum on the third interval [2, 3] is 20 at x=2.
Minimum on the fourth interval [3, 4] is 30 at x=3.
Therefore, the sum of areas of all rectangles equals to
S2 = 0·1 + 10·1 +
+ 20·1 + 30·1 = 60
.

To proceed to a general case, assume we have divided our segment into N equal parts, and find the limit of the area of all rectangles as N→∞.
In this case the width of each interval equals to
Δxi = 4/N
Right margin of i-th interval is
xi = i·4/N
The function value at this right margin is
f(xi) = 10·i·4/N
The area of the i-th rectangle, constructed on the i-th interval as a base and having height calculated above, equals to
Si = 10·i·(4/N)·(4/N) =
= (160/N²)·i

Our task is to sum these areas for i changing from 1 to N and find the limit of this sum as N→∞.
Summation by i is simple, we did this before (see "Algebra - Sequence and Series" chapter of this course or prove the following by induction).
Recall that
Σ[1,K]i = K·(K+1)/2

Therefore, the result of summation of the areas of N rectangles is
Σ[1,N]Si=(160/N²)·N·(N+1)/2=
= 80·(1+(1/N))

As N→∞, this value converges to 80, which constitutes the "area under curve".

Incidentally, our "area under curve" can be calculated as the area of a right triangle with base 4 and height 10·4=40, which equals to S=(1/2)·4·40=80 - the same answer as we have received through rather complicated summation and taking the limit.
That confirms the validity of our answer.

The end.
________________

Example 2

Find "area under curve" for f(x) = −x2+1 on segment [a=−1, b=1].

Solution

Again, let's consider two particular cases prior to generalize the problem. We will use the values of a function on the right margin of each interval.

Case 1. N=2
Point of division in two equal parts is x1=0
So, a=x0=−1x1=0x2=1=b
The function value on the right margin of the first interval [−1, 0] is 1 at x=0.
The function value on the right margin of the second interval [−1, 0] is 0 at x=1.
Therefore, the sum of areas of all rectangles equals to
S1 = 1·1 + 0·1 = 1.

Case 2. N=4
Two additional points of division in four equal parts are x1=−0.5 and x3=0.5
So, a=x0=−1x1=−0.5x2=0,
x3=0.5x4=1=b
The function value on the right margin of the first interval [−1, −0.5] is 3/4 at x=−0.5.
The function value on the right margin of the second interval [−0.5, 0] is 1 at x=0.
The function value on the right margin of the third interval [0, 0.5] is 3/4 at x=0.5.
The function value on the right margin of the fourth interval [0.5, 1] is 0 at x=1.
Therefore, the sum of areas of all rectangles equals to
S2 = (3/4)·0.5 + 1·0.5 +
+ (3/4)·0.5 + 0·0.5 = 5/4
.

To proceed to a general case, assume we have divided our segment into N equal parts, and find the limit of the area of all rectangles as N→∞.
In this case the width of each interval equals to
Δxi = 2/N
Right margin of i-th interval is
xi = −1+i·2/N
The function value at this right margin is
f(xi) = 1−(−1+i·2/N)² =
= (4/N)·i−(4/N²)·i²

The area of the i-th rectangle, constructed on the i-th interval as a base and having height calculated above, equals to
Si = Δxi·f(xi) =
= (2/N)·
[(4/N)·i−(4/N²)·i²] =
= (8/N²)·i−(8/N³)·i²

Our task is to sum these areas for i changing from 1 to N and find the limit of this sum as N→∞.
Recall that
Σ[1,K]i = K·(K+1)/2
Σ[1,K]i² = K·(K+1)·(2K+1)/6

Therefore, the result of summation of the areas of N rectangles is
Σ[1,N]Si = (8/N²)·(N·(N+1)/2 −
− (8/N³)·N·(N+1)·(2N+1)/6

As N→∞, this value converges to 4−(16/6)=4/3, which constitutes the "area under curve".

The end.