The average value of a continuous function f(x) over an interval [a, b] is
[tex]\displaystyle f_{\mathrm{ave}[a,b]} = \frac1{b-a}\int_a^b f(x)\,dx[/tex]
We're given that
[tex]\displaystyle f_{\rm ave[-1,2]} = \frac13 \int_{-1}^2 f(x) \, dx = -4[/tex]
[tex]\displaystyle f_{\rm ave[2,7]} = \frac15 \int_2^7 f(x) \, dx = 8[/tex]
and we want to determine
[tex]\displaystyle f_{\rm ave[-1,7]} = \frac18 \int_{-1}^7 f(x) \, dx[/tex]
By the additive property of definite integration, we have
[tex]\displaystyle \int_{-1}^7 f(x) \, dx = \int_{-1}^2 f(x)\,dx + \int_2^7 f(x)\,dx[/tex]
so it follows that
[tex]\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(\int_{-1}^2 f(x)\,dx + \int_2^7 f(x)\,dx\right)[/tex]
[tex]\displaystyle f_{\rm ave[-1,7]} = \frac18 \left(3\times(-4) + 5\times8\right)[/tex]
[tex]\displaystyle f_{\rm ave[-1,7]} = \boxed{\frac72}[/tex]
