Unit 6.13 – Evaluating Improper Integrals

AP® Calculus BC ONLY | Integrals with Infinite Limits or Discontinuities

Why This Matters: Improper integrals extend integration to cases where either the interval is infinite or the integrand has a discontinuity! These integrals arise naturally in probability, physics, and engineering. An improper integral may converge (have a finite value) or diverge (infinite or undefined). This BC-only topic uses limits to determine convergence. Mastering improper integrals is essential—they appear on virtually every BC exam and test your understanding of both integration and limits!

🎯 What Are Improper Integrals?

DEFINITION

An integral is improper if one or both of the following occurs:

Type 1: One or both limits of integration are infinite
Type 2: The integrand has a discontinuity (usually infinite) at or between the limits

📝 Key Idea: We evaluate improper integrals using limits to determine if they converge to a finite value or diverge.

📈 Type 1: Infinite Limits of Integration

Type 1 Improper Integrals

Case 1: Upper limit is ∞

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

Case 2: Lower limit is −∞

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

Case 3: Both limits are infinite

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

(Split at any convenient point \(c\), often \(c = 0\))

Convergence vs. Divergence:

✓ CONVERGES if:

The limit exists and equals a finite number

\[ \lim_{b \to \infty} \int_a^b f(x)\,dx = L \quad \text{(finite)} \]

✗ DIVERGES if:

The limit is infinite or does not exist

\[ \lim_{b \to \infty} \int_a^b f(x)\,dx = \pm\infty \text{ or DNE} \]

⚠️ Type 2: Discontinuous Integrand

Type 2 Improper Integrals

Case 1: Discontinuity at lower limit \(a\)

\[ \int_a^b f(x) \, dx = \lim_{t \to a^+} \int_t^b f(x) \, dx \]

Case 2: Discontinuity at upper limit \(b\)

\[ \int_a^b f(x) \, dx = \lim_{t \to b^-} \int_a^t f(x) \, dx \]

Case 3: Discontinuity at interior point \(c\), where \(a < c < b\)

\[ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \]

\[ = \lim_{t \to c^-} \int_a^t f(x)\,dx + \lim_{s \to c^+} \int_s^b f(x)\,dx \]

📖 Worked Examples: Type 1 (Infinite Limits)

Example 1: Convergent Integral with Upper Limit ∞

Problem: Evaluate \(\int_1^{\infty} \frac{1}{x^2} \, dx\)

Solution:

Step 1: Set up as a limit

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

Step 2: Evaluate the definite integral

\[ = \lim_{b \to \infty} \int_1^b x^{-2} \, dx = \lim_{b \to \infty} \left[-x^{-1}\right]_1^b \]

\[ = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} - (-\frac{1}{1})\right) \]

\[ = \lim_{b \to \infty} \left(1 - \frac{1}{b}\right) \]

Step 3: Evaluate the limit

\[ = 1 - 0 = 1 \]

Answer: The integral CONVERGES to 1

Example 2: Divergent Integral

Problem: Evaluate \(\int_1^{\infty} \frac{1}{x} \, dx\)

Solution:

Step 1-2: Set up and integrate

\[ \int_1^{\infty} \frac{1}{x} \, dx = \lim_{b \to \infty} \int_1^b \frac{1}{x} \, dx \]

\[ = \lim_{b \to \infty} [\ln|x|]_1^b = \lim_{b \to \infty} (\ln b - \ln 1) \]

\[ = \lim_{b \to \infty} \ln b \]

Step 3: Evaluate the limit

\[ = \infty \]

Answer: The integral DIVERGES (equals ∞)

Example 3: Both Limits Infinite

Problem: Evaluate \(\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx\)

Solution:

Step 1: Split at \(x = 0\)

\[ \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \int_{-\infty}^0 \frac{1}{1+x^2}\,dx + \int_0^{\infty} \frac{1}{1+x^2}\,dx \]

Step 2: Evaluate first integral

\[ \int_{-\infty}^0 \frac{1}{1+x^2}\,dx = \lim_{a \to -\infty} [\arctan x]_a^0 \]

\[ = \lim_{a \to -\infty} (\arctan 0 - \arctan a) = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} \]

Step 3: Evaluate second integral

\[ \int_0^{\infty} \frac{1}{1+x^2}\,dx = \lim_{b \to \infty} [\arctan x]_0^b \]

\[ = \lim_{b \to \infty} (\arctan b - \arctan 0) = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

Step 4: Add both parts

\[ \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

Answer: The integral CONVERGES to \(\pi\)

📖 Worked Examples: Type 2 (Discontinuous Integrand)

Example 4: Discontinuity at Lower Limit

Problem: Evaluate \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)

Solution:

Identify discontinuity: \(\frac{1}{\sqrt{x}}\) is undefined at \(x = 0\)

Step 1: Set up as limit (from right)

\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^1 \frac{1}{\sqrt{x}} \, dx \]

Step 2: Integrate

\[ = \lim_{t \to 0^+} \int_t^1 x^{-1/2} \, dx = \lim_{t \to 0^+} [2x^{1/2}]_t^1 \]

\[ = \lim_{t \to 0^+} (2\sqrt{1} - 2\sqrt{t}) = \lim_{t \to 0^+} (2 - 2\sqrt{t}) \]

Step 3: Evaluate limit

\[ = 2 - 0 = 2 \]

Answer: The integral CONVERGES to 2

Example 5: Discontinuity at Upper Limit

Problem: Evaluate \(\int_0^1 \frac{1}{(1-x)^2} \, dx\)

Solution:

Identify: Discontinuity at \(x = 1\) (upper limit)

Step 1: Set up limit (from left)

\[ \int_0^1 \frac{1}{(1-x)^2} \, dx = \lim_{t \to 1^-} \int_0^t \frac{1}{(1-x)^2} \, dx \]

Step 2: Integrate (use substitution \(u = 1-x\))

\[ = \lim_{t \to 1^-} \left[\frac{1}{1-x}\right]_0^t = \lim_{t \to 1^-} \left(\frac{1}{1-t} - \frac{1}{1-0}\right) \]

\[ = \lim_{t \to 1^-} \left(\frac{1}{1-t} - 1\right) \]

Step 3: Evaluate limit

As \(t \to 1^-\), \(1-t \to 0^+\), so \(\frac{1}{1-t} \to +\infty\)

Answer: The integral DIVERGES

Example 6: Discontinuity at Interior Point

Problem: Evaluate \(\int_0^2 \frac{1}{x-1} \, dx\)

Solution:

Identify: Discontinuity at \(x = 1\) (interior point)

Step 1: Split at discontinuity

\[ \int_0^2 \frac{1}{x-1}\,dx = \int_0^1 \frac{1}{x-1}\,dx + \int_1^2 \frac{1}{x-1}\,dx \]

Step 2: Evaluate first integral

\[ \int_0^1 \frac{1}{x-1}\,dx = \lim_{t \to 1^-} [\ln|x-1|]_0^t = \lim_{t \to 1^-} (\ln|t-1| - \ln 1) \]

As \(t \to 1^-\), \(|t-1| \to 0^+\), so \(\ln|t-1| \to -\infty\)

Answer: The integral DIVERGES (first part diverges, so whole integral diverges)

🔬 The p-Integral Test

p-Integral Test (Important for BC!)

For Type 1 (infinite upper limit):

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

Converges if \(p > 1\)
Diverges if \(p \leq 1\)

For Type 2 (discontinuity at 0):

\[ \int_0^1 \frac{1}{x^p} \, dx \]

Converges if \(p < 1\)
Diverges if \(p \geq 1\)

📝 Memorize This! These results are extremely useful for quick answers and comparison tests!

💡 Essential Tips & Strategies

✅ General Strategy:

Step 1: Identify the type (infinite limits OR discontinuity)
Step 2: Set up appropriate limit(s)
Step 3: Evaluate the definite integral
Step 4: Evaluate the limit(s)
Step 5: Conclude convergence or divergence

🔥 Quick Checks:

Exponential decay: \(\int_a^{\infty} e^{-kx}\,dx\) converges for \(k > 0\)
Polynomial denominator: Higher power in denominator → more likely to converge
p-integral test: Use it to quickly determine convergence
Both infinite limits: MUST evaluate both separately
Interior discontinuity: BOTH sides must converge for whole integral to converge

Decision Tree:

Is there infinity in the limits?

YES: Type 1 → Set up limit as \(b \to \infty\) or \(a \to -\infty\)
NO: Check for discontinuities...

Is the integrand discontinuous in \([a,b]\)?

At \(a\): Type 2 → Limit as \(t \to a^+\)
At \(b\): Type 2 → Limit as \(t \to b^-\)
Interior point \(c\): Split and do both sides
NO: Regular definite integral!

❌ Common Mistakes to Avoid

Mistake 1: Not recognizing the integral is improper
Mistake 2: Evaluating \(\int_1^{\infty} f(x)\,dx\) as \(F(\infty) - F(1)\) without using limits
Mistake 3: For both infinite limits, not splitting and evaluating separately
Mistake 4: Missing discontinuities (always check if integrand is defined on interval!)
Mistake 5: Wrong limit direction (should be \(t \to a^+\) for discontinuity at \(a\))
Mistake 6: Concluding convergence when one part diverges (ALL parts must converge)
Mistake 7: Sign errors when evaluating limits
Mistake 8: Forgetting absolute value in \(\ln|x-a|\)
Mistake 9: Arithmetic errors with infinity (\(\infty - \infty\) is NOT 0!)
Mistake 10: Not stating final conclusion (converges/diverges)

📝 Practice Problems

Set A: Type 1 (Infinite Limits)

\(\int_1^{\infty} \frac{1}{x^3} \, dx\)
\(\int_2^{\infty} \frac{1}{x\ln x} \, dx\)
\(\int_0^{\infty} e^{-2x} \, dx\)

Answers:

Converges to \(\frac{1}{2}\)
Diverges (limit is \(\infty\))
Converges to \(\frac{1}{2}\)

Set B: Type 2 (Discontinuous Integrand)

\(\int_0^1 \frac{1}{x^{1/3}} \, dx\)
\(\int_0^1 \frac{1}{x} \, dx\)
\(\int_{-1}^1 \frac{1}{x^2} \, dx\)

Answers:

Converges to \(\frac{3}{2}\) (since \(p = \frac{1}{3} < 1\))
Diverges (since \(p = 1\))
Diverges (discontinuity at \(x = 0\), interior point)

✏️ AP® Exam Success Tips

What AP® BC Graders Look For:

Recognize improper integral: State why it's improper
Set up limit correctly: Write the limit notation explicitly
Show integration: Don't skip the antiderivative step
Evaluate at bounds: Show the substitution
Evaluate the limit: Show limit evaluation
State conclusion: "Converges to ___" or "Diverges"
For split integrals: Evaluate both parts separately

⚡ Ultimate Quick Reference

IMPROPER INTEGRALS CHECKLIST

Complete Reference Guide
Type	Setup	Notes
Upper limit ∞	\(\lim_{b \to \infty} \int_a^b f(x)\,dx\)	Most common type
Lower limit −∞	\(\lim_{a \to -\infty} \int_a^b f(x)\,dx\)	Less common
Both limits infinite	Split at \(c\), evaluate both	Both must converge
Discontinuity at \(a\)	\(\lim_{t \to a^+} \int_t^b f(x)\,dx\)	Approach from right
Discontinuity at \(b\)	\(\lim_{t \to b^-} \int_a^t f(x)\,dx\)	Approach from left
Interior discontinuity	Split and do both sides	Both must converge

p-Integral Test:

\(\int_1^{\infty} \frac{1}{x^p}\,dx\): Converges if \(p > 1\)
\(\int_0^1 \frac{1}{x^p}\,dx\): Converges if \(p < 1\)

Master Improper Integrals! An integral is improper if limits are infinite (Type 1) or integrand has discontinuity (Type 2). Type 1: \(\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx\). Converges if limit is finite, diverges otherwise. For both infinite limits, MUST split and evaluate separately—both must converge. Type 2: If discontinuity at \(a\), use \(\lim_{t \to a^+} \int_t^b f(x)\,dx\); at \(b\), use \(\lim_{t \to b^-} \int_a^t f(x)\,dx\); interior point, split. p-Integral Test: \(\int_1^{\infty} \frac{1}{x^p}\,dx\) converges iff \(p > 1\); \(\int_0^1 \frac{1}{x^p}\,dx\) converges iff \(p < 1\). Always: (1) identify type, (2) set up limit, (3) integrate, (4) evaluate limit, (5) state conclusion. Common functions: \(\int_0^{\infty} e^{-kx}\,dx\) converges for \(k > 0\). Check BOTH parts when split—if either diverges, whole integral diverges. Show all limit notation clearly on exams! This BC-only topic appears frequently—practice until automatic! 🎯✨