Find the average value and the value(s) of c guaranteed by the mean value theorem for integrals if f left parenthesis x right parenthesis equals 3 x squared plus 2 x over the interval left square bracket 0 comma 2 right square bracket.A. f subscript a v g end subscript equals 12c equals fraction numerator short dash 1 plus square root of 37 over denominator 3 end fractionB. f subscript a v g end subscript equals 6c equals fraction numerator short dash 1 minus square root of 19 over denominator 3 end fraction space a n d space c equals fraction numerator short dash 1 plus square root of 19 over denominator 3 end fractionC. f subscript a v g end subscript equals 12c equals fraction numerator short dash 1 minus square root of 37 over denominator 3 end fraction space a n d space c equals fraction numerator short dash 1 plus square root of 37 over denominator 3 end fractionD. f subscript a v g end subscript equals 6c equals fraction numerator short dash 1 plus square root of 19 over denominator 3 end fraction

The question is asking for the average value of the function f(x) = 3x^2 + 2x over the interval [0,2] and the value(s) of c guaranteed by the Mean Value Theorem for integrals.

Step 1: Find the average value of the function over the interval [0,2]. The average value of a function f(x) over the interval [a,b] is given by the formula:

f_avg = (1/(b-a)) * ∫ from a to b of f(x) dx

In this case, a = 0, b = 2, and f(x) = 3x^2 + 2x. So we have:

f_avg = (1/2) * ∫ from 0 to 2 of (3x^2 + 2x) dx

To evaluate the integral, we use the power rule for integration, which states that the integral of x^n dx is (1/(n+1))x^(n+1). Applying this rule, we get:

∫ from 0 to 2 of (3x^2 + 2x) dx = [x^3 + x^2] from 0 to 2 = 8 + 4 = 12

So f_avg = (1/2) * 12 = 6.

Step 2: Find the value(s) of c guaranteed by the Mean Value Theorem for integrals. The Mean Value Theorem for integrals states that there exists at least one number c in the interval [a,b] such that f(c) = f_avg. In other words, we need to solve the equation 3c^2 + 2c = 6 for c.

This is a quadratic equation, which we can solve using the quadratic formula c = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = 3, b = 2, and c = -6, so we have:

c = [-2 ± sqrt((2)^2 - 43(-6))] / (23) c = [-2 ± sqrt(4 + 72)] / 6 c = [-2 ± sqrt(76)] / 6 c = [-2 ± 2sqrt(19)] / 6 c = [-1 ± sqrt(19)] / 3

So the correct answer is B. f_avg = 6, c = (-1 - sqrt(19))/3 and c = (-1 + sqrt(19))/3.