# Definite Integral:

Definite integral is a form of Integral or Anti derivative in which we don’t get a range of answer or indefinite answer , Instead we get a fixed or definite answer.

Or, A definite integral is the integral of a function in a closed interval and it is denoted by: $\int^b_a f(x) . d(x)$

Which means , the Integral or Anti derivative of the function “f(x)” in an interval from “a” to “b”.

# Definite Integral Formula:

We can calculate or evaluate a definite integral using the definite integral formula which states that:

If “f” is continuous on [a,b] and $\phi$ is any antiderivative of “f” then , $\int^b_a f(x) . dx = \phi (b) - \phi (a)$

Now let us prove the formula:

We can prove the formula using the fundamental theorem of calculus which states that:

If “f” is continuous function and $F(x) = \int^x_a f(t) .dt$  , then , $\dfrac{d}{x} F(x) = f(x)$

Now let us prove the definite integral formula with the help of fundamental theorem of calculus,

Proof:

Let, $F(x) = \int^x_a f(t) .dt$

As “a” is a constant obviously we get “F(a) = 0″.

And as, “F” and $\phi$ are antiderivatives of same function “f” , they differ by a constant “c”  as stated in antiderivatives .

So, $F(x) = \phi (x) + c$ For some constant “c”.

Thus , $F(a) = \phi (a) + c$

or, $0 = \phi (x) + c$

or , $\phi (x) = - c$

So , $F(b) = \phi (b) - \phi (a)$

And we also have, $F(b) = \int^b_a f(t) .dt$

Thus , $\int^b_a f(t) . dt = \phi (b) - \phi (a)$

# Example of Definite Integrals:

We can evaluate definite integrals using the same techniques of integration we used while evaluating indefinite integrals.

One example of definite integral is: $\int^3_0 x^5 .dx$

We can easily evaluate this integral using Integration Formulas .

So we get: $\int^3_0 x^5 .dx = \left( \dfrac{1}{6} x^6 \right)^3_0$ $= \frac{1}{6} ( 3^6 - 0^6)$ $= \frac{243}{2}$

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