Calculate the area bounded by the curve y = 1/x

Calculate the area bounded by the curve $y = \frac{1}{x}$, the lines $x = 1$, $x = a$, and $y = 0$, where $a > 0$. What does $P(a)$ approach as $a \to 0$?

We need to consider two cases depending on whether $a < 1$ or $a > 1$.

Case 1: $a > 1$

The region is between $x = 1$ and $x = a$ where $a > 1$.

Step 1: Set up the integral

\[P(a) = \int_1^a \frac{1}{x} \, dx\]

Step 2: Evaluate

\[P(a) = [\ln|x|]_1^a = \ln a - \ln 1 = \ln a - 0 = \ln a\]

\[\boxed{P(a) = \ln a \quad \text{for } a > 1}\]

Case 2: $0 < a < 1$

The region is between $x = a$ and $x = 1$ where $0 < a < 1$.

Step 1: Set up the integral

\[P(a) = \int_a^1 \frac{1}{x} \, dx\]

Step 2: Evaluate

\[P(a) = [\ln|x|]_a^1 = \ln 1 - \ln a = 0 - \ln a = -\ln a\]

Since $0 < a < 1$, we have $\ln a < 0$, so $-\ln a > 0$.

We can also write:

\[-\ln a = \ln\frac{1}{a} = \ln a^{-1}\]

\[\boxed{P(a) = -\ln a = \ln\frac{1}{a} \quad \text{for } 0 < a < 1}\]

Unified formula

We can write a unified formula valid for all $a > 0$, $a \neq 1$:

\[\boxed{P(a) = |\ln a|}\]

Or more explicitly:

\[P(a) = \begin{cases} \ln a & \text{if } a > 1 \\ -\ln a = \ln\frac{1}{a} & \text{if } 0 < a < 1 \end{cases}\]

Behavior as $a \to 0$

For the case $0 < a < 1$, as $a \to 0^+$:

\[P(a) = -\ln a = \ln\frac{1}{a}\]

Since $\frac{1}{a} \to +\infty$ as $a \to 0^+$:

\[\lim_{a \to 0^+} P(a) = \lim_{a \to 0^+} \ln\frac{1}{a} = \lim_{a \to 0^+} (-\ln a) = +\infty\]

\[\boxed{\lim_{a \to 0^+} P(a) = +\infty}\]

Interpretation

As $a \to 0^+$, the interval $[a, 1]$ extends further to the left, and the area under the curve $y = \frac{1}{x}$ grows without bound because:

• The function $\frac{1}{x} \to +\infty$ as $x \to 0^+$
• The integral $\int_a^1 \frac{1}{x} \, dx$ diverges to $+\infty$ as $a \to 0^+$

Graphical understanding

• For $a = 0.1$: $P(0.1) = -\ln(0.1) = \ln 10 \approx 2.303$
• For $a = 0.01$: $P(0.01) = -\ln(0.01) = \ln 100 \approx 4.605$
• For $a = 0.001$: $P(0.001) = -\ln(0.001) = \ln 1000 \approx 6.908$

The area increases without bound as $a$ approaches 0.

Summary

Area formula:

\[\boxed{P(a) = \begin{cases} \ln a & \text{if } a > 1 \\ -\ln a = \ln\frac{1}{a} & \text{if } 0 < a < 1 \end{cases}}\]

Limit as $a \to 0^+$:

\[\boxed{\lim_{a \to 0^+} P(a) = +\infty}\]

Key insights:

1. For $a > 1$: area between $x = 1$ and $x = a$ is $P(a) = \ln a$
2. For $0 < a < 1$: area between $x = a$ and $x = 1$ is $P(a) = -\ln a$
3. As $a \to 0^+$: the area grows to infinity
4. The hyperbola $y = \frac{1}{x}$ has unbounded area near $x = 0$
5. This illustrates that $\int_0^1 \frac{dx}{x}$ is an improper integral that diverges