Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 79

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 functionf(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.