The fundamental theorem of calculus is approximately named because it establish a connection between the two branches of calculus, differential calculus and integral calculus. That is fundamental theorem of calculus gives the precise inverse relationship between derivative and the integral.

It is crucial that we understand the distinction between a definite integral and an indefinite integral. The fundamental theorem will help us to learn more about this.

Fundamental theorem of calculus, which is the central theorem of integral calculus. It connects integration and differentiation, enabling us to compute integrals using an anti derivative of the integrand function rather than by taking limits of Riemann sums. Leibniz and Newton exploited this relationship and started mathematical development that fueled the scientific revolution for the next 200 years.
If $f(x)$ is continues at every point of [a, b], and $F(x)$ is an antiderivative of $f(x)$ on [a,b], then

$\int_{a}^{b} f(x)dx$ = $F(b)-F(a)$
If $f(x)$ is continuous on [a,b], then the derivative of the function $F(x)$ = $\int_{a}^{x} f(t)dt$ is:

$\frac{dF}{dx}$ = $\frac{d}{dx}$$\int_{a}^{x}f(t)dt$ = $f(x)$
Let us begin by assuming that $f$ is a non negative and continuous on an interval [a,b].

Proof of Fundamental Theorem of Calculus
The area A under the graph of $f$ over the interval [a,b] is represented by the definite integral

$A$ =$\int _{a}^{b}f(x)dx$

If $A(x)$ is the area under the graph of $f$ from $a$ to $x$
,
$A'(x)$ = $f(x)$, it states that $A(x)$ is a anti derivative of $f(x)$, which implies that every other antiderivative of $f(x)$ on [a,b] can be obtained by adding a constant to $A(x)$.

Accordingly let $F(x)$ =$A(x)+C$ be any antiderivative of $f(x)$, and when we substrate $F(a)$ from $F(b)$,

$F(b)-F(a)$ = $[A(b)+C] – [A(a)+C]$

= $A(b) -A(a)$

= $\int_{a}^{b}f(x)dx - \int_{a}^{a}f(x)dx$

= $\int_{a}^{b}f(x)dx – 0$

= $\int_{a}^{b}f(x)dx$

hence we have the expression

$\int_{a}^{b}f(x)dx$ = $F(b)-F(a)$

The theorem says that to calculate the definite integral of $f$ over [a,b] all we need to do is:
1. Find the anti derivative $F$ of $f$, and
2. calculate the number $\int_{a}^{b}f(x)dx$ = $F(b)-F(a)$.
The usual notation for $F(b)-F(a)$ is $[F(x)]_{a}^{b}$ depending on whether $F$ has one or more terms.


Proof of second Fundamental Theorem:
If $f(t)$ is an integrable function over a finite interval $I$, then the integral from any fixed number $a \in I$ to another number $x \in I$ defines a new function $F$ whose value at $x$ is

$F(x)$ = $\int _{a}^{x}f(t)dt$

The above equation defines a way to define new functions, but its importance now is the connection it makes between integrals and derivatives.

We prove the fundamental theorem by applying the definition of the derivative directly to the function $F(x)$, when $x$ and $x+h$ are in [a, b]. this means writing out the difference quotient

$\frac{F(x+h)-F(x)}{h}$...........(1)

and showing that its limit as $h \to 0$ is the number $f(x)$ for each $x$ in [a, b].
When we replace $F(x+h)$ and $F(x)$ by their defining integrals, we can write

$F(x+h)-F(x)$ = $\int _{a}^{x+h}f(t)dt-\int _{a}^{x}f(t)dt$

by simplifying the right hand side by adding rule for integrals we will have

$\int _{x}^{x+h}f(t)dt$

so that the equation (1) become

$\frac{F(x+h)-F(x)}{h}$ = $\frac{1}{h}$$[F(x+h)-F(x)]$

= $\frac{1}{h}$$\int _{x}^{x+h}f(t)dt$

According to mean value theorem for definite integrals, the value of the last expression in the above equation is one of the values taken on by $f$ in the interval between $x$ and $x+h$, that is for some number $c$ in this interval ,

$\frac{1}{h}$$\int _{x}^{x+h}f(t)dt$ = $f(c)$

As $h \to 0$,$x+h$ approaches $x$, forcing $c$ to approach $x$. since $f$ is continuous at $x$, $f(c)$ approaches $f(x)$:

$\lim_{h \to 0}f(c)$ = $f(x)$

from the definition of the derivative we have

$\frac{dF}{dx}$ = $\lim_{h \to 0}$$\frac{F(x+h)-F(x)}{h}$

= $\lim_{h \to 0}$$\int _{x}^{x+h}f(t)dt$

= $\lim_{h \to 0}$$f(c)$

= $f(x)$

Hence we have

$\frac{dF}{dx}$ = $\frac{d}{dx}$$\int_{a}^{x}f(t)dt$ = $f(x)$
The following are the examples of fundamental theorem of calculus.

Solved Examples

Question 1: Evaluate $\int_{1}^{2} xdx$.
Solution:
 
Let $f(x)$ = $x$ be the given function.

The function $F(x)$ = $\frac{1}{2}x^{2}$ is the antiderivative of $f(x)$ = $x$

thus $\int_{1}^{2} xdx$ = $\frac{1}{2}$$x^{2}]_{1}^{2}$

= $\frac{1}{2}$$(2)^{2}$ - $\frac{1}{2}$$(1)^{2}$

= $2-$$\frac{1}{2}$

= $\frac{3}{2}$
 

Question 2: Find $\frac{d}{dx}$$\int_{2}^{x}(1-t^{3})dt$
Solution:
 
Make use the second fundamental theorem to solve this problem

$\frac{d}{dx}$$\int_{2}^{x}(1-t^{3})dt$ = $\frac{d}{dx}$$[t-$$\frac{t^{4}}{4}$$]_{2}^{x}$

= $\frac{d}{dx}$$[(x-\frac{x^{4}}{4})-(2-\frac{2^{4}}{4})]$

= $\frac{d}{dx}$$[x-$$\frac{x^{4}}{4}$$+2]$

= $(1- x^{4})$