RaySix

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Created by @nathanedwards

at November 1st 2023, 3:35:22 am.

AP_Calculus_AB-Questions

#integrals

#average-value

#mean-value-theorem

Question:

Let the function f be defined by

f(x) = √(x^2 + 1) - 3x + 1.

(a) Find the average value of f(x) on the interval [0, 2].

(b) Determine all values of c in the interval (0, 2) guaranteed by the Mean Value Theorem for Integrals.

Answer:

(a) To find the average value of a function on an interval, we need to compute the definite integral of the function over that interval and then divide by the width of the interval. For this problem, the average value of f(x) on the interval [0, 2] can be calculated using the following formula:

$\text{Average\ Value}\ =\ \frac{1}{b-a}\int_{a}^{b}f(x)dx$

Substituting the given values into the formula, we have:

$\text{Average\ Value}\ =\ \frac{1}{2-0}\int_{0}^{2}\left(\sqrt{x^2+1}-3x+1\right)dx$

Calculating the integral using antiderivatives:

$\text{Average\ Value}\ =\ \frac{1}{2}\left.\left( \frac{1}{2}x\sqrt{x^2+1}-\frac{3}{2}x^2+x\right)\right|_{0}^{2}$

Evaluating the antiderivative expression at the upper and lower limits:

$\text{Average\ Value}\ =\ \frac{1}{2}\left[ \left( \frac{1}{2}(2)\sqrt{(2)^2+1}-\frac{3}{2}(2)^2+2\right) - \left( \frac{1}{2}(0)\sqrt{(0)^2+1}-\frac{3}{2}(0)^2+0\right) \right]$

Simplifying the expression:

$\text{Average\ Value}\ =\ \frac{1}{2}\left[ \frac{1}{2}(2)\sqrt{(5)}-\frac{3}{2}(4)+2 - 0 \right]$ $\text{Average\ Value}\ =\ \frac{1}{2}\left[ \sqrt{(5)}-6+2 \right]$

Calculating the expression:

$\text{Average\ Value}\ =\ \frac{1}{2}( \sqrt{5}-4 )$

Therefore, the average value of f(x) on the interval [0, 2] is <img src="https://latex.codecogs.com/svg.latex?\frac{1}{2}( \sqrt{5}-4 )" title="\frac{1}{2}( \sqrt{5}-4 )".

(b) According to the Mean Value Theorem for Integrals, if a function f(x) is continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that:

$f'(c)\ =\ \frac{f(b)-f(a)}{b-a}$

In this case, the interval is (0, 2), and the function f(x) = √(x^2 + 1) - 3x + 1 is continuous on [0, 2] and differentiable on (0, 2).

To find the values of c as guaranteed by the Mean Value Theorem for Integrals, we need to calculate f'(x) and then solve the equation:

$\frac{f(2)-f(0)}{2-0}\ =\ f'(c)$

Taking the derivative of f(x):

$f'(x)\ =\ \left(\sqrt{x^2+1}\right)'-(3x)'+(1)'$

Using the chain rule and power rule:

$f'(x)\ =\ \frac{1}{2}\cdot\frac{1}{\sqrt{x^2+1}}\cdot(2x)-3+0$

Simplifying the derivative:

$f'(x)\ =\ \frac{x}{\sqrt{x^2+1}}-3$

Now, substitute the values into the Mean Value Theorem equation:

$\frac{f(2)-f(0)}{2-0}\ =\ \frac{2}{\sqrt{2^2+1}}-3$

Calculating the expression:

$\frac{f(2)-f(0)}{2-0}\ =\ \frac{2}{\sqrt{5}}-3$

Simplifying the expression:

$\frac{f(2)-f(0)}{2-0}\ =\ \frac{2}{\sqrt{5}}-3$

Since the equation is

$\frac{2}{\sqrt{5}}-3\ =\ \frac{2}{\sqrt{5}}-3$

...it implies that any x value from the open interval (0, 2) satisfies the equation. Therefore, all values of c in the interval (0, 2) are guaranteed by the Mean Value Theorem for Integrals.

Hence, the values of c that satisfy the Mean Value Theorem for Integrals in the interval (0, 2) are any real number between 0 and 2.

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Mean Value Theorem for Integrals on [0, 2]