Consider integral from 0 to ³ dx/((x-2)²). Obviously

Consider $\int_0^3 \frac{dx}{(x-2)^2}$. Obviously $\int \frac{dx}{(x-2)^2} = -\frac{1}{x-2}$, so $\int_0^3 \frac{dx}{(x-2)^2} = -\left.\frac{1}{x-2}\right|_0^3 = -\frac{3}{2}$. But the integrand $\frac{1}{(x-2)^2}$ is positive in the given interval, so the integral over the given limits must be positive. Where is the error?

This is an important example that illustrates a common error when dealing with improper integrals.

Identifying the error

The error: The fundamental mistake is treating this as a proper definite integral when it is actually an improper integral.

Why is this an improper integral?

The integrand $f(x) = \frac{1}{(x-2)^2}$ has a vertical asymptote (discontinuity) at $x = 2$, which lies inside the integration interval $[0, 3]$.

At $x = 2$:

\[\lim_{x \to 2} \frac{1}{(x-2)^2} = +\infty\]

Therefore, the function is not continuous on $[0, 3]$, and we cannot directly apply the Newton-Leibniz formula:

\[\int_0^3 \frac{dx}{(x-2)^2} \neq \left[-\frac{1}{x-2}\right]_0^3\]

This direct application is invalid!

Correct approach: Improper integral

Since the integrand has a discontinuity at $x = 2 \in (0, 3)$, we must split the integral and take limits:

\[\int_0^3 \frac{dx}{(x-2)^2} = \lim_{\epsilon_1 \to 0^+} \int_0^{2-\epsilon_1} \frac{dx}{(x-2)^2} + \lim_{\epsilon_2 \to 0^+} \int_{2+\epsilon_2}^3 \frac{dx}{(x-2)^2}\]

Evaluate the first part:

\[\int_0^{2-\epsilon_1} \frac{dx}{(x-2)^2} = \left[-\frac{1}{x-2}\right]_0^{2-\epsilon_1}\]

\[= -\frac{1}{(2-\epsilon_1)-2} - \left(-\frac{1}{0-2}\right)\]

\[= -\frac{1}{-\epsilon_1} + \frac{1}{2} = \frac{1}{\epsilon_1} + \frac{1}{2}\]

Taking the limit:

\[\lim_{\epsilon_1 \to 0^+} \left(\frac{1}{\epsilon_1} + \frac{1}{2}\right) = +\infty\]

Evaluate the second part:

\[\int_{2+\epsilon_2}^3 \frac{dx}{(x-2)^2} = \left[-\frac{1}{x-2}\right]_{2+\epsilon_2}^3\]

\[= -\frac{1}{3-2} - \left(-\frac{1}{(2+\epsilon_2)-2}\right)\]

\[= -1 + \frac{1}{\epsilon_2}\]

Taking the limit:

\[\lim_{\epsilon_2 \to 0^+} \left(-1 + \frac{1}{\epsilon_2}\right) = +\infty\]

Conclusion:

\[\int_0^3 \frac{dx}{(x-2)^2} = +\infty + (+\infty) = +\infty\]

The integral diverges (does not exist as a finite number).

Summary of the error

1. What went wrong: The Newton-Leibniz formula was applied without checking whether the integrand is continuous on $[0, 3]$.
2. The discontinuity: The function $\frac{1}{(x-2)^2}$ is undefined (infinite) at $x = 2$.
3. Correct interpretation: Since the integrand is positive and approaches infinity near $x = 2$, the integral must diverge to $+\infty$, which is consistent with the geometric interpretation (infinite area).
4. The "answer" $-\frac{3}{2}$: This nonsensical negative result came from mechanically applying a formula without verifying its preconditions.

General lesson

Before applying the Newton-Leibniz formula $\int_a^b f(x) \, dx = F(b) - F(a)$, always check:

• Is $f(x)$ continuous on $[a, b]$?
• Are there any points where $f(x)$ is undefined or infinite?
• Are the limits $a$ and $b$ finite?

If any of these conditions fail, you are dealing with an improper integral, which requires limits.

Final answer:

\[\boxed{\int_0^3 \frac{dx}{(x-2)^2} = +\infty \text{ (diverges)}}\]

The error was applying the Newton-Leibniz formula to a function with a discontinuity inside the integration interval without using the proper limit definition for improper integrals.