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Basic Concepts Of Definite Integral Part 2

Posted on - 04-01-2017

Math

IIT JEE

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 .

=

=

= = .

 
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