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Rzqust [24]
3 years ago
11

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

Mathematics
1 answer:
barxatty [35]3 years ago
7 0

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.

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