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.
No comments:
Post a Comment