Recall the **First Fundamental Theorem of Calculus:**

If f is continuous on closed interval [a,b], we follow:

`int_a^bf(x)dx` = F(b) - F(a)

where F is the anti-derivative of f on [a,b].

This shows that we need to solve first the indefinite integral F(x) to be able to apply the difference of values F based on the given boundary limit of a and b.

The resulting value will be the definite integral.

For the given problem `int_(-1)^(2)2^xdx` , the integrand function`f(x) = 2^x`

which is in a form of a exponential function.

The basic integration formula for exponential function follows:

`int a^u du = a^u/(ln(a))`

By comparison: `a^u` vs `2^x` , we may let:

`a=2` , `u=x` and then` du= dx`

Then applying the formula, we get:

`int 2^x dx = 2^x/(ln(2))`

indefinite integral function` F(x) = 2^x/(ln(2))`

Applying the formula:` int_a^(b) f(x) dx = F(b)-F(a)` :

Based on the given problem: `int_(-1)^(2)2^x dx` , the boundary limits are:

lower limit:`a= -1` and upper limit:`b = 2`

Plug-in the boundary limits in ` F(x) =2^x/(ln(2)) ` one at a time, we get:

`F(a) = F(-1)= (2^(-1))/ln(2)`

`F(a) F(-1)=1/(2ln(2))`

`F(b) =F(2)= 2^2/(ln(2))`

`F(b)=4/(ln(2))`

Solving for the definite integral:

`F(b)-F(a) = F(2) - F(-1)`

`= 4 /(ln(2)) - 1/(2ln(2))`

` = 4 *1/(ln(2)) -(1/2)*1/(ln(2))`

` = (4 - 1/2) *1/(ln(2))`

`= 7/2*1/(ln(2))`

or `7/(2ln(2)) ` as the** Final Answer**.