We have to find the definite integral of y=3^4x for x=0 to x=1

Now Int y = Int [ 3^4x]

= 3^4x / 4* ln 3

Now for x = 1.

3^4x / 4* ln 3 = 3^4/4* ln 3

For x = 0

3^4x / 4* ln 3 = 3^0 / 4*ln 3

=> 1 / 4* ln 3

Subtracting the values for x = 1 and x = 0

=> 3^4/4* ln 3 - 1 / 4* ln 3

=> 80 / 4* ln 3

=> 20 / ln 3

**Therefore the required result is 20/ ln 3**

Find the value of definite integral y=3^4x for x=0 to x=1.

F(x) = Int y dx = Int3^(4x) dx.

We put 4x = t.

Then dx = dt/4.

Therefore Int 3^4xdx = Int 3^t *dt/4 = 3^t/4ln3, as Int a^x dx = a^x/lna

Therefore Int 3^4xdx = 3^4x/4log3.

Therefore F(x) = 3^4x/4ln3.

Therefore the area under y = 3^4x fron x = 0 to x= 1, is F(1)-F(0) = {3^4*1 - 3^4*0}/4ln3 = 80/4ln3

Therefore the area under y = 3^4x from x= 0 to x= 1 is 80/4ln3.

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 **