Question

Find the numbers b such that the average value of f(x) = 7 + 10x − 9x2 on the interval [0, b] is equal to 8.

Answer 1

Answer:

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$.

Step-by-step explanation:

The mean value of function within a given interval is given by the following integral:

$\bar f = \frac{1}{b-a}\cdot \int\limits^b_a {f(x)} \, dx$

If $f(x) = 7 +10\cdot x - 9\cdot x^{2}$, $a = 0$, $b = b$ and $\bar f = 8$, then:

$\frac{1}{b}\cdot \int\limits^b_0 {7+10\cdot x -9\cdot x^{2}} \, dx = 8$

$\frac{7}{b}\int\limits^b_0 \, dx + \frac{10}{b} \int\limits^b_0 {x}\, dx - \frac{9}{b} \int\limits^b_0 {x^{2}}\, dx = 8$

$\left(\frac{7}{b} \right)\cdot b + \left(\frac{10}{b} \right)\cdot \left(\frac{b^{2}}{2} \right)-\left(\frac{9}{b} \right)\cdot \left(\frac{b^{3}}{3} \right) = 8$

$7 + 5\cdot b - 3\cdot b^{2} = 8$

$3\cdot b^{2}-5\cdot b +1 = 0$

The roots of this polynomial are determined by the Quadratic Formula:

$b_{1} \approx 1.434$ and $b_{2} \approx 0.232$.

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$.