Question

Find a and b such that f(x)=x^2+ax+b has an average value of 34/3 over [0,1] and an average value of 88/3 over [0,5]?

Answer 1

I got `a = 37` and `b = -15/2`

Explanation:

Recall that average value is given by

`A = 1/(b - a) int_a^b f(x) dx`

Thus

`34/3 = int_0^1 x^2 + ax + b dx`

`34/3 = [1/3x^3 + a/2x^2 + bx]_0^1`

`34/3 = 1/3 + 1/2a + b`

`11 = 1/2a + b`

`22 = a + 2b`

The second equation will be

`88/3 = 1/(5 - 0) int_0^5 f(x) dx`

`440/3 = int_0^5 f(x) dx`

`440/3 = [1/3x^3 + a/2x^2 + bx]_0^5`

`440/3 = 1/3(5)^3 + 25/2a + 5b`

`105 = 25/2a + 5b`

`210 = 25a + 10b`

We can now readily solve our system `{(210 = 25a + 10b), (22 = a + 2b):}` using substitution or elimination.

`210 = 25(22 - 2b) + 10b`

`210 = 510 - 50b + 10b`

`b = -7.5`
Since

`a = 22 - 2b`, we know that `a = 22 - 2(-7.5) = 37`

Hopefully this helps!