Calculus II: Fundamental Theorem of Calculus

The definite integral of a function is defined as a limit of Riemann sum. It is the area under the graph of this function between two limits. The Riemann sun method used to evaluate a definite integral is long and some times difficult or tedious. Isaac Newton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) discovered a simple method to evaluate definite integrals. This method is known as The Fundamental theorem of Calculus. This theorem is based on the link between derivative and primitive or anti-derivative. It gives the value of the definite integral of a function as a difference between anti-derivative of the function at the upper and lower limits of integration.

1. Properties of definite integral

From the Riemann sum:

b n

∫ f(x) dx = lim Σ f(ci) Δxi

a n → ∞ i=1

and letting that all the involved integrals exist, that is the involved functions are derivable on the related interval [a, b], we have the following properties:

1.1. Properties1:
Inverting the bounds of the integral

If we sum from b to a, the width of each sub-interval is negative in the the interval [a, b], with a> b. That is:

a n

∫ f(x) dx = lim Σ f(ci) (- Δxi)

b n → ∞ i=1

That is

a n

∫ f(x) dx = - lim Σ f(ci) Δxi

b n → ∞ i=1

Therefore

b a

∫ f(x) dx = - ∫ f(x) dx

a b

1.2. Properties2:
Linearity of the integral

The constant c ∊ R

b b

∫ c f(x) dx = c ∫ f(x) dx

a a

1.3. Properties13:
Adding & subtracting integrals

b b b

∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx

a a a

1.4. Properties4:
Integrals via a point in the interval

The constants a, b, and c ∊ R

b c b

∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx

a a c

This formula is valid for a< c < b or a < b < c.

All the above formulas are proved easily by using the sum and the limit properties in the Riemann sum.

2.Mean Value Theorem for Integrals

Let f a function continuous on the interval [a, b]. The mean value M of this function f on the interval [a, b] is given by the formula:

   1 b

M =
∫ f(x) dx

b - a a

If we subdivide the interval [a, b] in n equal sub-intervals :

x0= a, x1, x2, ..., xn= b

The mean value M_n is written:

M_n = (f(x1) + f(x2) + f(x3) + ... + f(xn))/n

With Δxi = xi - xi-1 = (b - a)/n, we have:

M_n = (1/(b - a)) (f(x1) + f(x2) + f(x3) + f(xn))Δxi

The exact value of M_n is M.

M = lim M_n =
     n → ∞

   1 n    1 b

lim
Σ f(xi) Δxi =
∫ f(x) dx

n→ ∞ b - a i = 1 b - a a

This formula is written:

b

M (b - a) = ∫ f(x) dx

a

M if the value of the function of a real c between a and b in the interval [a, b]. The above formula can be then written as:

b

∫ f(x) dx = (b - a) f(c)

a

3.Fundamental theorem of Calculus

Now we are interested to find the relationship between Riemann sum, definite integral, and primitive of a function f continuous on an interval [a, b]. In other word, If F(x) is a primitive of f(x), what is the link between F(x) and

b

∫ f(x) dx

a

This integral of f(x) from the lower bound a to the upper bound b is the definition or the notation of the Riemann sum

b n

∫ f(x) dx = limit Σ f(ci) Δxi

a n → ∞ i=1

We want to establish the relationship between the integral and the derivative, the primitive, and no longer use Riemann sum, and therefore find a rapid way to evaluate a definite integral by using only the primitive. The related formula is often called Newton-Leibniz formula or the fundamental theorem of Calculus.

To prove this theorem, we use two parts. The first part shows the link between the integral and the primitive, and the second part shows the way to evaluate an integral of a function, by using its primitive.

3.1. Part 1: Primitive of a function

The definite integral of a continuous function f(t) in the interval [a, b] exists and it is is areal value. It is equal to

b

∫ f(x) dx

a

Now, if the upper bound varies as x the long of the interval [a, b], (x∊ [a, b]) the definite integral it becomes:

x

∫ f(x) dx

a

Therefore, the integral varies with respect to the variable x.

Let F(x) this function, we have

x

F(x) = ∫ f(t) dt

a

and determine to what it corresponds.

Increasing x by Δx, the above statement becomes:

x+Δx

F(x+Δx) = ∫ (f(t) dt

a

The difference between the two above expresion is:

x+Δx x

F(x+Δx) - F(x) = ∫ (f(t) dt - ∫ f(t) dt

a a

Since

x+Δx x x+Δx

∫ f(t) dt = ∫ f(t) dt + ∫ f(t) dt

a a x

we have

x+Δx

F(x+Δx) - F(x) = ∫ (f(t) dt

x

Using the mean value theorem for integrals, we obtain

F(x+Δx) - F(x) = Δx f(c).

The real c is located between x and x+Δx (x∊[x, x+Δx])

Dividing by Δx

(F(x+Δx) - F(x))/ Δx = f(c)

And taking the limits

lim (F(x+Δx) - F(x))/Δx = lim f(c)

Δx→0 c → x

When Δx → 0 (Δx + x - x)→ 0 then [x, x+Δx] → [x].
As c ∊[x, x+Δx], c → x.

The above formula becomes

F'(x) = f(x)

That is: F(x) is a primitive of f(x).

x

F(x): a primitive of f(x) = ∫ f(t) dt

a

t is a dummy variable .

3.2. Part 2: Primitive of a function

If F(x) is a primitive of the function f(x), so is

x

∫ f(t) dt .

a

The primitives of a function are different from each other just by a constant. So

x

∫ f(t) dt = F(x) + const.

a

This relationship is valid for any x on the interval [a, b], then true for x = b, and for x = a. Therefore

For x = b

b

∫ f(t) dt = F(b) + const.

a

For x = a

a

∫ f(t) dt = 0 = F(a) + const.

a

Hence

const = - F(a)

Finally,

b

∫ f(t) dt = F(b) - F(a)

a

The variable of integration is dummy, let use x instead:

b

∫ f(x) dx = F(b) - F(a)

a

This is the fundamental theorem of Calculus, known as the Newton-Leibniz formula.

4. Example

f(x) = 1/√x.

First: find a primitive:

∫ f(x) dx = 2 √x

Second: Make sure that the primitive is continuous in the considered interval [1, 9]. It is.

Third:
Applying the fundamental theorem of calculus, we have immediately the result:

9 9

∫ f(x) dx = [2 √x] = 6 - 2 = 4

1 1