Find the average value of the given function f(x) over the specified interval a < x <b. Round your answer to the third decimal place. f(x) = x + 2/x^2 + 4x + 5 over -1 < x < 1 V=? Note: < this has an equal sign on bottom

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The average value of a function f(x) on the interval [a,b] is given by `1/(b-a)int_a^bf(x)dx` :

Then the average value of `f(x)=(x+2)/(x^2+4x+5)` on [-1,1] is:

`1/(1-(-1))int_(-1)^1 (x+2)/(x^2+4x+5)dx=1/2int_(-1)^1(x+2)/(x^2+4x+5)dx`

If we multiply the integral by `1/2` and the integrand by 2 we get:

`=1/4int_(-1)^1(2x+4)/(x^2+4x+5)dx` Let `u=x^2+4x+5,du=2x+4 dx`

If x=-1 then u=2, if...

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The average value of a function f(x) on the interval [a,b] is given by `1/(b-a)int_a^bf(x)dx` :

Then the average value of `f(x)=(x+2)/(x^2+4x+5)` on [-1,1] is:

`1/(1-(-1))int_(-1)^1 (x+2)/(x^2+4x+5)dx=1/2int_(-1)^1(x+2)/(x^2+4x+5)dx`

If we multiply the integral by `1/2` and the integrand by 2 we get:

`=1/4int_(-1)^1(2x+4)/(x^2+4x+5)dx` Let `u=x^2+4x+5,du=2x+4 dx`

If x=-1 then u=2, if x=1 then u=10 so we have:

`=1/4int_2^(10)u^(-1)du=1/4[ln|u| |_2^(10)]`

`=1/4[ln|10|-ln|2|]`

`=1/4ln5`

`~~.402`

The average value of f(x) on theindicated interval is approximately .402

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