Math / Calculus

QuestionFind the average value of the function $f$ over the interval $[0,4]$ . $f(x)=\frac{12}{x+1}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the average value of a function over a specific range.
Think of it like finding the average height of a curvy mountain range between two points. Watch out! Don't just plug in the endpoints!
We need the average value, not just the value at the ends.

STEP 2

1. Set up the average value formula
2. Calculate the definite integral
3. Divide by the length of the interval

STEP 3

The average value of a function $f(x)$ over an interval $[a, b]$ is given by: $\frac{1}{b - a} \int_{a}^{b} f(x) \, dx$ This formula basically says, "find the area under the curve, and then divide by the width of the interval to get the average height." So cool!

STEP 4

In our case, $f(x) = \frac{12}{x + 1}$ , $a = \textbf{0}$ , and $b = \textbf{4}$ .
Let's plug those values into our formula: $\frac{1}{4 - 0} \int_{0}^{4} \frac{12}{x + 1} \, dx$ $\frac{1}{4} \int_{0}^{4} \frac{12}{x + 1} \, dx$

STEP 5

We can pull the constant 12 out of the integral to make things easier: $\frac{12}{4} \int_{0}^{4} \frac{1}{x + 1} \, dx$ $3 \int_{0}^{4} \frac{1}{x + 1} \, dx$

STEP 6

Now, we integrate!
The integral of $\frac{1}{x + 1}$ is $\ln|x + 1|$ .
Don't forget the absolute value, but in this case, since our interval is $[0, 4]$ , $x + 1$ is always positive anyway! $3 \Big[ \ln|x + 1| \Big]_{0}^{4}$

STEP 7

Now, we plug in our limits of integration, 4 and 0: $3 \Big[ \ln(4 + 1) - \ln(0 + 1) \Big]$ $3 \Big[ \ln(5) - \ln(1) \Big]$ Since $\ln(1) = 0$ , we have: $3 \ln(5)$

STEP 8

We already divided by the length of the interval (which was 4) when we simplified $\frac{12}{4}$ to 3 in the beginning.
So, we're already here!

STEP 9

The average value of the function $f(x) = \frac{12}{x + 1}$ over the interval $[0, 4]$ is $3 \ln(5)$ .

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