Question Find the average value of $f(x) = 2x$ on the interval $[-1, 1]$ .

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = 2x$ .
2. We are looking for the average value of $f(x)$ on the interval $[-1, 1]$ .

STEP 2

The average value of a continuous function $f(x)$ on the interval $[a, b]$ is given by the formula:
$\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx$

STEP 3

Identify the interval limits $a$ and $b$ for the given function. In this case, $a = -1$ and $b = 1$ .

STEP 4

Substitute the values of $a$ and $b$ into the average value formula.
$\text{Average value} = \frac{1}{1 - (-1)} \int_{-1}^{1} 2x \, dx$

STEP 5

Simplify the denominator of the fraction.
$\text{Average value} = \frac{1}{1 + 1} \int_{-1}^{1} 2x \, dx$
$\text{Average value} = \frac{1}{2} \int_{-1}^{1} 2x \, dx$

STEP 6

Calculate the integral of $2x$ from $-1$ to $1$ .
$\int_{-1}^{1} 2x \, dx = \left[ x^2 \right]_{-1}^{1}$

STEP 7

Evaluate the integral at the upper and lower limits.
$\left[ x^2 \right]_{-1}^{1} = (1)^2 - (-1)^2$

STEP 8

Simplify the result of the evaluation.
$\left[ x^2 \right]_{-1}^{1} = 1 - 1$
$\left[ x^2 \right]_{-1}^{1} = 0$

STEP 9

Now, multiply the result of the integral by the fraction outside the integral to find the average value.
$\text{Average value} = \frac{1}{2} \cdot 0$

STEP 10

Calculate the average value.
$\text{Average value} = 0$
The average value of $f(x) = 2x$ on $[-1, 1]$ is 0, which corresponds to option b. None.

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