Find the average value of the function on the given

Find the average value of the function on the given interval: $$ y = \ln x, \quad [1, e] $$

Definition of average value

The average value of a function $f(x)$ on the interval $[a, b]$ is:

$$\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx$$

Step 1: Set up the formula

For $f(x) = \ln x$ on $[1, e]$:

$$\bar{y} = \frac{1}{e-1} \int_1^e \ln x \, dx$$

Step 2: Calculate $\int_1^e \ln x \, dx$

We use integration by parts.

Let $u = \ln x$ and $dv = dx$.

Then: $du = \frac{1}{x} \, dx$ and $v = x$

By integration by parts:

$$\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} \, dx = x\ln x - \int 1 \, dx$$

$$= x\ln x - x + C = x(\ln x - 1) + C$$

Step 3: Evaluate the definite integral

$$\int_1^e \ln x \, dx = [x(\ln x - 1)]_1^e$$

At $x = e$:

$$e(\ln e - 1) = e(1 - 1) = e \cdot 0 = 0$$

Wait, let me recalculate more carefully:

$$e(\ln e - 1) = e(1 - 1) = 0$$

At $x = 1$:

$$1(\ln 1 - 1) = 1(0 - 1) = -1$$

Therefore:

$$\int_1^e \ln x \, dx = 0 - (-1) = 1$$

Step 4: Calculate the average value

$$\bar{y} = \frac{1}{e-1} \int_1^e \ln x \, dx = \frac{1}{e-1} \cdot 1 = \frac{1}{e-1}$$

$$\boxed{\bar{y} = \frac{1}{e-1}}$$

Step 5: Numerical value

$$\bar{y} = \frac{1}{e-1} \approx \frac{1}{2.71828 - 1} = \frac{1}{1.71828} \approx 0.582$$

$$\boxed{\bar{y} \approx 0.582}$$

Verification

Let's verify our calculation of the integral.

$$\int_1^e \ln x \, dx = [x\ln x - x]_1^e$$

At $x = e$:

$$e \ln e - e = e \cdot 1 - e = e - e = 0$$

At $x = 1$:

$$1 \ln 1 - 1 = 1 \cdot 0 - 1 = -1$$

Therefore:

$$\int_1^e \ln x \, dx = 0 - (-1) = 1 \quad \checkmark$$

Interpretation

The average value $\bar{y} = \frac{1}{e-1} \approx 0.582$ represents the height of a rectangle with base $[1, e]$ that has the same area as the region under the curve $y = \ln x$ from $x = 1$ to $x = e$.

Check reasonableness:

• At $x = 1$: $\ln 1 = 0$
• At $x = e$: $\ln e = 1$

The average should be somewhere between 0 and 1, and indeed $0.582 \in (0, 1)$. ✓

Moreover, since $\ln x$ is concave on this interval, we expect the average to be less than the midpoint value. At $x = \frac{1+e}{2} \approx 1.859$:

$$\ln(1.859) \approx 0.620$$

Our average of $0.582$ is slightly less, which makes sense for a concave function. ✓

Summary

Average value of $y = \ln x$ on $[1, e]$:

$$\boxed{\bar{y} = \frac{1}{e-1} \approx 0.582}$$

Key steps:

1. Use formula: $\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx$
2. Integration by parts: $\int \ln x \, dx = x\ln x - x + C$
3. Evaluate: $\int_1^e \ln x \, dx = [x\ln x - x]_1^e = 0 - (-1) = 1$
4. Calculate average: $\bar{y} = \frac{1}{e-1} \cdot 1 = \frac{1}{e-1}$

Key insights:

• Integration by parts is essential for $\int \ln x \, dx$
• The antiderivative is $x(\ln x - 1)$
• At $x = e$, the expression $x\ln x - x$ simplifies nicely to 0
• The average value lies between the minimum (0) and maximum (1)
• The exact answer $\frac{1}{e-1}$ is a beautiful closed form