How do you use the integral fomula to find the average value of the function #f(x)=18x# over the interval between 0 and 4?

Question

Aurora Alvarado · Accepted Answer

I found: #y_(av)=36#

Consider a general case:

Evelyn Agard · Answer

undefined To find the average value of the function $ f(x) = 18x $ over the interval [0, 4], you can use the formula for the average value of a function on a closed interval $[a, b]$:

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

So, for $ f(x) = 18x $ over the interval [0, 4], the average value would be:

$$ \text{Average value} = \frac{1}{4 - 0} \int_{0}^{4} 18x \, dx $$

$$ = \frac{1}{4} \left[ \frac{18x^2}{2} \right]_{0}^{4} $$

$$ = \frac{1}{4} \left[ \frac{18(4)^2}{2} - \frac{18(0)^2}{2} \right] $$

$$ = \frac{1}{4} \left[ \frac{18 \times 16}{2} - 0 \right] $$

$$ = \frac{1}{4} \times 144 $$

$$ = 36 $$

So, the average value of $ f(x) = 18x $ over the interval [0, 4] is 36.

William Abdalla · Answer

undefined To find the average value of the function $ f(x) = 18x $ over the interval $[0, 4]$, you can use the integral formula:

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

where $ a = 0 $ and $ b = 4 $ in this case. Plugging in the values:

$$
\text{Average value} = \frac{1}{4-0} \int_{0}^{4} 18x \, dx
$$

Now, integrate $ 18x $ with respect to $ x $:

$$
\int 18x \, dx = 9x^2 + C
$$

Then, evaluate the integral from $ 0 $ to $ 4 $:

$$
\int_{0}^{4} 18x \, dx = \left[9x^2\right]_{0}^{4} = 9(4)^2 - 9(0)^2 = 9(16) - 9(0) = 144
$$

Substitute this value back into the formula for average value:

$$
\text{Average value} = \frac{1}{4-0} \times 144 = \frac{144}{4} = 36
$$

So, the average value of the function $ f(x) = 18x $ over the interval $[0, 4]$ is $ 36 $.