First
Fundamental Theorem of Calculus
If f (x) is a
continuous function on [a, b], then F (x) = 
f (t) dt is
differentiable at every point x in [a, b] and 
F (x) = 
" x ∈ (a, b). This is called the first
fundamental theorem of calculus.
Existence of
Anti–derivative of Continuous Functions:
If y = f (x) is continuous on [a, b], then
there exists a function F (x) whose derivative on [a, b] is f. i.e. every
continuous function is the derivative of some of the functions. In other words,
every continuous function has an anti derivative. However, not every anti
derivative even when it exists, is expressible, in closed form, in terms of
elementary functions. e.g. 
In
all such cases, the anti derivative is obviously some new function which does
not reduce to a combination of a finite number of elementary functions.
Second Fundamental Theorem
of Calculus
If f(x) is a
continuous function on [a, b] and F(x) is any anti derivative of f(x) on [a, b]
i.e. 
(x) = f (x) " x ∈ (a, b) , then 
(also called the
Newton-Leibnitz formula).
The function F(x) is the integral of
f(x) and a and b are the lower and the upper limits of integration.
Proof:
From the first
fundamental theorem 
i.e., the expression
within the bracket must be constant in the interval and hence we can write 
, where c is some real
constant.
Thus 
Hence, 
The second
fundamental theorem is used to calculate the value of the definite integral. A
note of caution to you is that f (t) must be continuous in [a, b] or else you
will have to partition it into subintervals such that f(x) is continuous in
each of the subintervals. .
Change of Variables in
Definite Integration
If the functions
f(x) is continuous on [a, b] and the function x = g(t) is continuously differentiable
on the interval [t1, t2] and a = g (t1) and b
= g (t2), then
.
Improper Integrals
Another
type of improper integral is that in which the integrand is not defined for one
or both limits of integration. Here we take limiting values as shown in the
following example.
Example.1
Evaluate
.
Solution
Note that ln x is
not defined for x = 0. Hence we can write .
I = 
dx
= 
(Integrating by
parts.)
= 
=
Now, 
(0 x (- ¥) form)
= 
= 
(using L’ Hospital
Rule)
= 
= 0
Hence, I = –1
Example.2
Prove
that 
.
Solution
Let I = 
Put x +
&⇒ 
or dx = 
Also 
so that dx = 
.
When x 
0, t 1 and
when x 
, t .
= 
= 
= 
= 
.
