To find the average value of a function, use the formula `1/{b-a}int_a^b f(t)dt`.

This means that the average of `cos t` in an interval is:

`1/{b-a}int_a^b cos t dt`

`=1/{b-a} (sin b-sin a)`

`={sin b-sin a}/{b-a}`

To answer Part A of the question, we use that result for the intervals:

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To find the average value of a function, use the formula `1/{b-a}int_a^b f(t)dt`.

This means that the average of `cos t` in an interval is:

`1/{b-a}int_a^b cos t dt`

`=1/{b-a} (sin b-sin a)`

`={sin b-sin a}/{b-a}`

To answer Part A of the question, we use that result for the intervals:

`[0,pi]`** has average `1/pi(sin pi-sin 0) = 0` .**

**`[0,pi/2]` has average `1/{pi/2}(sin (pi/2)-sin 0) = 2/pi` .**

**`[0,pi/4]` has average `1/{pi/4}(sin (pi/4)-sin0)=4/{pi sqrt 2}` .**

**`[0,0.01]` has average `1/0.01(sin 0.01-sin0)=0.99998` .**