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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 $b_{2} \approx 0.232$ .

7 0

3 years ago

Which line has the smallest slope? y= 3x v=9x y= .4x y= .25x

pentagon [3]

Y=.25x is the answer

3 0

3 years ago

Read 2 more answers

Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5

lakkis [162]

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer: v = 5v + 4u + 1.5sin(3t),

0

1

4

5

0

1.5sin(3t)

1

2.5

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t) where u'(1) = 2.5 u(1) = 1

v represents the "velocity function" i.e v = u'(t)

As v = u'(t)

u' = v

since u' = v

v' = u"

v' = 5u' + 4u + 1.5sin(3t) ( given that u" - 5u' - 4u = 1.5sin(3t) )

= 5v + 4u + 1.5sin(3t) ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ = $\left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]$ $\left[\begin{array}{ccc}u&\\v&\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]$

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

$\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]$ = $\left[\begin{array}{ccc}1\\2.5\\\end{array}\right]$

7 0

3 years ago