Question

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`

Answer 1

`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 `t`, thus:

`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`