Answer:
The numbers [tex]b[/tex] such that the average value of [tex]f(x) = 7 +10\cdot x - 9\cdot x^{2}[/tex] on the interval [0, b] is equal to 8 are [tex]b_{1} \approx 1.434[/tex] and [tex]b_{2} \approx 0.232[/tex].
Step-by-step explanation:
The mean value of function within a given interval is given by the following integral:
[tex]\bar f = \frac{1}{b-a}\cdot \int\limits^b_a {f(x)} \, dx[/tex]
If [tex]f(x) = 7 +10\cdot x - 9\cdot x^{2}[/tex], [tex]a = 0[/tex], [tex]b = b[/tex] and [tex]\bar f = 8[/tex], then:
[tex]\frac{1}{b}\cdot \int\limits^b_0 {7+10\cdot x -9\cdot x^{2}} \, dx = 8[/tex]
[tex]\frac{7}{b}\int\limits^b_0 \, dx + \frac{10}{b} \int\limits^b_0 {x}\, dx - \frac{9}{b} \int\limits^b_0 {x^{2}}\, dx = 8[/tex]
[tex]\left(\frac{7}{b} \right)\cdot b + \left(\frac{10}{b} \right)\cdot \left(\frac{b^{2}}{2} \right)-\left(\frac{9}{b} \right)\cdot \left(\frac{b^{3}}{3} \right) = 8[/tex]
[tex]7 + 5\cdot b - 3\cdot b^{2} = 8[/tex]
[tex]3\cdot b^{2}-5\cdot b +1 = 0[/tex]
The roots of this polynomial are determined by the Quadratic Formula:
[tex]b_{1} \approx 1.434[/tex] and [tex]b_{2} \approx 0.232[/tex].
The numbers [tex]b[/tex] such that the average value of [tex]f(x) = 7 +10\cdot x - 9\cdot x^{2}[/tex] on the interval [0, b] is equal to 8 are [tex]b_{1} \approx 1.434[/tex] and [tex]b_{2} \approx 0.232[/tex].
