The average value of an integrable function on an interval [a,b] is defined as:
`"Avg"= 1/(b-a) int_(a)^(b) f(x)dx `
Find the average value of `f(t)= t sin(t^2) dt ` on [0,10]:
`"Avg"= 1/10 int_0^10 t sin(t^2) dt `
Let `u=t^2, du=2tdt ==> tdt=du/2 ` ; also x=0 ==> u=0 and...
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The average value of an integrable function on an interval [a,b] is defined as:
`"Avg"= 1/(b-a) int_(a)^(b) f(x)dx `
Find the average value of `f(t)= t sin(t^2) dt ` on [0,10]:
`"Avg"= 1/10 int_0^10 t sin(t^2) dt `
Let `u=t^2, du=2tdt ==> tdt=du/2 ` ; also x=0 ==> u=0 and x=10 ==> u=100 so we can rewrite as:
` =1/10 int_0^100 sin(u) 1/2 du ` or factoring out the constant:
`=1/20 int_0^100 sin(u)du `
`=1/20[ -cos(u) |_0^100 ] `
`=1/20[-cos(100)+cos(0)] `
`~~.0069 `
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The average value of f(t) on [0,10] is approximately .0069
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