Topic 6.13 Notes – Evaluating Improper Integrals

What an Improper Integral Is

A definite integral is improper if you cannot evaluate it directly with the Fundamental Theorem of Calculus because:

1. One or both bounds are infinite
   $$\int_3^{\infty} f(x)\,dx, \quad \int_{-\infty}^2 f(x)\,dx, \quad \int_{-\infty}^{\infty} f(x)\,dx$$
2. The integrand is unbounded on the interval
   (There’s a vertical asymptote at an endpoint or inside.)

Example:

$$\int_0^2 \frac{1}{\sqrt{x}}\,dx$$

The function blows up at $x=0$, so this is improper.

Here’s the visual idea. The graph of $y = \frac{1}{\sqrt{x}}$ has a vertical asymptote at $x=0$. The curve shoots upward near that value, but we can still ask whether the shaded area from 0 to 2 settles to a finite number.

Improper integral with a vertical asymptote at $x=0$

The big question every time is: Does the area settle to a finite number?

Rewriting as a Limit

Limits are what make this possible. You replace the “infinite” or undefined part with a variable and take a limit.

Infinite Upper Bound

$$\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$$

Infinite Lower Bound

$$\int_{-\infty}^{b} f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx$$

Both Bounds Infinite

Split it at any convenient number $c$:

$$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_{c}^{\infty} f(x)\,dx$$

Each part must converge. If either diverges, the whole thing diverges.

Vertical Asymptote at an Endpoint

If the issue is at $x=0$:

$$\int_0^4 f(x)\,dx = \lim_{a \to 0^+} \int_a^4 f(x)\,dx$$

Vertical Asymptote Inside the Interval

If there’s an asymptote at $x=2$ in $[0,5]$:

$$\int_0^5 f(x)\,dx = \int_0^2 f(x)\,dx + \int_2^5 f(x)\,dx$$

Each piece gets its own limit.

Students often forget to split at interior asymptotes. On an FRQ, that costs points immediately.

Evaluating Step by Step

Once it’s written as a limit, everything else is normal calculus.

1. Rewrite as a limit.
2. Find the antiderivative.
3. Apply the Fundamental Theorem of Calculus.
4. Take the limit.
5. State clearly: converges to ___ or diverges.

Quick example:

$$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b\to\infty} \int_1^b x^{-2}\,dx$$

Antiderivative:
$$- x^{-1}$$

Evaluate:
$$\lim_{b\to\infty} \left(-\frac{1}{b} + 1\right)$$

As $b \to \infty$, $1/b \to 0$, so the result is $1$.
This integral converges.

p-Integrals You Must Know

These show up constantly and are the comparison model for many problems.

Case 1

$$\int_1^{\infty} \frac{1}{x^p}\,dx$$

• Converges if $p > 1$
• Diverges if $p \le 1$

Case 2

$$\int_0^1 \frac{1}{x^p}\,dx$$

• Converges if $p < 1$
• Diverges if $p \ge 1$

Notice how the rule flips depending on where the issue is. That detail is tested.

Behavior at Infinity

A fast-decaying function tends to converge.

• $e^{-x}$ shrinks very quickly → often converges.
• $1/x$ shrinks slowly → diverges.
• Higher powers like $1/x^3$ shrink faster → converge.

You’re judging how fast the function approaches zero as $x \to \infty$.

How It Appears on Tests

• “Evaluate or show that the integral diverges.”
• Partial fractions first, then apply limits.
• Substitution before evaluating the limit.
• Splitting at asymptotes.

On free-response, you must actually show the limit notation. Writing an antiderivative and plugging in infinity without the limit loses credit.