Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 80

To evaluate the integral: int_(-4)^(4) 3^(x/4) dx , we follow the formula based from the First Fundamental Theorem of Calculus:
int_a^bf(x)dx=F(b)- F(a)
wherein f is a continuous and F is the indefinite integral f on the closed interval [a,b].
Based on the given problem, the boundary limits are:
a =-4 and b=4
To solve for F as the indefinite integral of f, we follow the basic integration formula for an exponential function:
int a^u du = a^u/(ln(a))+C
By comparison: a^u vs 3^(x/4) , we let:
a=3 and u=x/4 then du= 1/4dx .
Rearrange du= 1/4dx into 4 du =dx .
Apply u-substitution using u =x/4 and 4du=dx :
int 3^(x/4) dx= int 3^u * 4du
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int 3^u * 4du =4 int 3^u du
Applying the formula: int a^u du = a^u/(ln(a)) .
4 int 3^u du= 4 *[ 3^u/(ln(3))]
Express in terms of x using u=x/4 :
4 *[ 3^u/(ln(3))] =4 *[ 3^(x/4)/(ln(3))]
Then indefinite integral function F(x) =4 *[ 3^(x/4)/(ln(3))]
Applying F(b) - F(a) with the closed interval [a,b] as [-4,4]:
int_(-4)^(4) 3^(x/4) dx =4 *[ 3^((4)/4)/(ln(3))] -4 *[ 3^(-4/4)/(ln(3))]
=(4*3^(1))/(ln(3)) - (4 * 3^(-1))/(ln(3))
=(12)/(ln(3))-4/(3ln(3))
= (12 -4/3) *1/(ln(3))
= ((36)/3-4/3)*1/(ln(3))
= (32/3)*1/(ln(3)) or (32)/(3ln(3)) as the Final Answer.