Question

Let f(x) = 3x(x – 2). For what value of c, to the nearest thousandth, is f(c) equal to the average value of f over the interval [-2, 2] –1. 517 –1. 236 –0. 915 –0. 528

Answer 1

Answer:

-1.517

Need to know:

Integral constant rule: ∫ (ax)dx = a ∫ (x)dx

Integral power rule: ∫ xⁿ = xⁿ⁺¹/(n + 1)

$\int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Quadratic formula: $\frac{-b +-\sqrt{b^{2} -4ac} }{2a}$

Step-by-step explanation:

First, we have to find the average value of f over the interval [-2, 2]

To find that, calculate the integral of 3x(x - 2)

∫ 3x(x - 2)dx = 3 ∫ x(x - 2)dx = 3 ∫ x² - 2xdx (pulled out the 3 because of the constant rule)

Use the power rule to solve the integral

3 [x²⁺¹/(2 + 1) - 2x¹⁺¹/(1 + 1)] = 3[x³/3 - 2x²/2]

Distribute the 3

3[x³/3 - 2x²/2] = x³ - 3x² = F(x)

We plug in 2 in place of x

F(2) = 2³ - 3(2)² = -4

Do the same with -2

F(-2) = (-2)³ - 3(-2)² = -20

We have to remember that  $\int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

let a = -2    b = 2

F(2) - F(-2) = -4 - (-20) = -4 + 20 = 16

The average value of f over the given interval is 16.

Now we make it to where f(x) = 16 to find c

3x(x - 2) = 16

3x² - 6x = 16

Subtract 16 from both sides to make it equal to zero

3x² - 6x = 16

       - 16    - 16

3x² - 6x - 16 = 0

a = 3 b = -6 c = 16

Plug these numbers into the quadratic formula

$\frac{-(-6) +-\sqrt{(-6)^{2} -4(3)(-16)} }{2(3)}$

$\frac{6 +-\sqrt{36 + 192} }{6}$

$\frac{6 +\sqrt{228} }{6}$ = -1.51661

$\frac{6 -\sqrt{228} }{6}$ = 3.51661

-1.51661 is the only number out of the two that fits in the given interval [-2,2], so c would have to be -1.517 in order for f(c) to equal the average value of f over the given interval.