If int_0^1 f(t) dt = 19 then find?: (A) int_0^0.125 f(8t) dt, (B) int_0^0.25 f(1−4t) dt, and (C) int_0.4^0.5 f(5-10t) dt
1 Answer
int_0^0.125 \ f(8t) \ dt = 19/8
int_0^0.25 \ f(1−4t) \ dt = 19/4
int_0.4^0.5 \ f(5-10t) \ dt = 9/10
Explanation:
We have:
int_0^1 \ f(t) \ dt = 19
We first note that the above result is independent of the variable
int_0^1 \ f(t) \ dt = int_0^1 \ f(u) \ du = 19
Part A
Find
I_1 = int_0^0.125 \ f(8t) \ dt
We can perform a substitution, Let:
u = 8t => (du)/(dt) = 8 => 1/8(du)/(dt) = 1
And we change the limits of integration:
t={ (0), (0.125) :} => u={ (0), (1) :}
If we substitute into the integral then we get:
I_1 = int_0^1 \ f(u) \ (1/8) \ du
\ \ \ = 1/8 \ int_0^1 \ f(u) \ du
\ \ \ = 1/8 * 19
\ \ \ = 19/8
Part B
Find
I_2 = int_0^0.25 \ f(1−4t) \ dt
We can perform a substitution, Let:
u = 1-4t => (du)/(dt) = -4 => -1/4(du)/(dt) = 1
And we change the limits of integration:
t={ (0), (0.25) :} => u={ (1), (0) :}
If we substitute into the integral then we get:
I_2 = int_1^0 \ f(u) \ (-1/4) \ du
\ \ \ = -1/4 \ int_1^0 \ f(u) \du
\ \ \ = 1/4 \ int_0^1 \ f(u) \du
\ \ \ = 1/4 * 19
\ \ \ = 19/4
Part C
Find
I_3 = int_0.4^0.5 \ f(5-10t) \ dt
We can perform a substitution, Let:
u = 5-10t => (du)/(dt) = -10 => -1/10(du)/(dt) = 1
And we change the limits of integration:
t={ (0.4), (0.5) :} => u={ (1), (0) :}
If we substitute into the integral then we get:
I_3 = int_1^0 \ f(u) \ (-1/10) \ du
\ \ \ = -1/10 \ int_1^0 \ f(u) du
\ \ \ = 1/10 \ int_0^1 \ f(u) du
\ \ \ = 1/10 * 19
\ \ \ = 19/10
