# Definite integral.How to find the value of definite integral of an exponential function, if the base is >1 and the limits of integration are 0 and 1?

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For f(x)=3^4x, to find the definite integral between x = 0 and x = 1, first find the indefinite integral.

F(0) = Int [ f(x) ] = 3^(4x) / 4*ln 3

The definite integral is F(1) - F(0)

=> 3^(4*1) / 4*ln 3 - 3^(4*0) / 4*ln 3

=> (1/4*ln 3)*80

=> 80/4*ln 3

=> 20/ln 3

I'll provide the following example.

First, I'll choose the exponential function f(x)=3^4x (notice that the base is 3>1).

To determine the value of the definite integral, we'll apply the Leibniz-Newton formula:

Int f(x)dx = F(b) - F(a), where x = a to x = b

We'll determine the indefinite integral:

Int (3^4x)dx

We'll substitute 4x = t.

For x = 0 => t = 0

For x = 1 => t = 4

We'll differentiate both sides:

4dx = dt

dx = dt/4

We'll re-write the integral in t:

(1/4)*Int (3^t)dt = (1/4)*(3^t)/ln 3

We'll calculate F(4) and F(0):

F(4) = (3^4)/ln 3^4

F(0) = 3^0/ln 3^4

F(4) - F(0) = (81 - 1)/ln 81

F(4) - F(0) = 80/ln 81

The value of definite integral Int (3^4x)dx, for x = 0 to x = 1, is:

**Int (3^4x)dx = 80/ln 81 **