Understanding Improper Integrals: Definitions and Practice

Arfa Momin

12/6/2025

AP Calculus AB/BC

Improper Integrals

•

Dec 6, 2025

•

3 pages

Understanding Improper Integrals: Definitions and Practice

Arfa Momin

@arfamomin_hkhz

Improper integrals help us find areas when dealing with infinity... Show more

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Improper Integrals
Q/: Can a shape with non-zero thickness and infinite length have a finite area?
Instructive Example : What is the aven un

Understanding Improper Integrals

Ever wonder if a shape with infinite length can have a finite area? That's what improper integrals help us figure out. When dealing with infinity, we can't just apply the Fundamental Theorem of Calculus directly.

Improper integrals of Type I handle infinite intervals. They're defined as:

For upper bound infinity: $\int_a^{\infty} f(x)dx = \lim_{t \to \infty} \int_a^t f(x)dx$
For lower bound negative infinity: $\int_{-\infty}^b f(x)dx = \lim_{t \to -\infty} \int_t^b f(x)dx$

When the limit exists, we say the integral is convergent, meaning the area is finite (for positive functions). If the limit doesn't exist, the integral is divergent.

💡 How quickly a function approaches zero matters! For example, $\int_1^{\infty} \frac{1}{x} dx$ diverges because it doesn't approach zero fast enough, but $\int_1^{\infty} \frac{1}{x^2} dx$ converges because it gets smaller quickly.

If both parts of an improper integral from negative infinity to positive infinity converge, we can add them together: $\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^a f(x) dx + \int_a^{\infty} f(x) dx$

Vertical Asymptotes and Type II Integrals

What happens when a function shoots up to infinity at a certain point? These vertical asymptotes create another type of improper integral. For example, $\frac{1}{\sqrt{x}}$ has a vertical asymptote at $x=0$ .

Type II improper integrals handle vertical asymptotes:

If function has a vertical asymptote at point $a$ : $\int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx$
If function has a vertical asymptote at point $b$ : $\int_a^b f(x) dx = \lim_{t \to b^-} \int_a^t f(x) dx$

For discontinuities within an interval, we break the integral at the problem point and evaluate each piece separately: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$

🔑 Remember these key examples: $\int_0^1 \frac{1}{x^p} dx$ and $\int_1^{\infty} \frac{1}{x^p} dx$ both converge when $p < 1$ and diverge when $p \geq 1$ . These "p-integrals" appear frequently on tests!

When evaluating these integrals, you'll need to compute the limits to justify whether they converge or diverge.

Evaluating Complex Improper Integrals

For general improper integrals, you'll need to break them down into simpler Type I and Type II pieces. This divide-and-conquer approach helps tackle complex problems systematically.

Remember this crucial rule: if any piece of the improper integral diverges, the entire integral diverges. For example, $\int_{-\infty}^{\infty} \frac{1}{x^2} dx$ must be split into parts to evaluate properly.

The Comparison Test is a powerful tool for determining convergence without calculating the exact value:

If $0 \leq g(x) \leq f(x)$ and $\int g(x)dx$ diverges → $\int f(x)dx$ also diverges
If $0 \leq g(x) \leq f(x)$ and $\int f(x)dx$ converges → $\int g(x)dx$ also converges

🧠 When facing complicated integrals, try bounding them with simpler functions whose convergence you already know. This strategy often saves you from difficult integrations!

For example, to evaluate $\int_{1}^{\infty} \frac{sin(x) + 2}{x^2} dx$ , notice that $0 \leq \frac{sin(x) + 2}{x^2} \leq \frac{3}{x^2}$ on $[1, \infty)$ . Since $\int_{1}^{\infty} \frac{3}{x^2} dx$ converges to 3, our original integral must also converge by the Comparison Test.

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AP Calculus AB/BC

•

Dec 6, 2025

•

3 pages

Understanding Improper Integrals: Definitions and Practice

Arfa Momin

@arfamomin_hkhz

Improper integrals help us find areas when dealing with infinity or vertical asymptotes. While regular integrals work with finite bounds and continuous functions, improper integrals tackle challenges like infinite intervals or points where functions become undefined.

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Understanding Improper Integrals

Improper integrals of Type I handle infinite intervals. They're defined as:

For upper bound infinity: $\int_a^{\infty} f(x)dx = \lim_{t \to \infty} \int_a^t f(x)dx$
For lower bound negative infinity: $\int_{-\infty}^b f(x)dx = \lim_{t \to -\infty} \int_t^b f(x)dx$

When the limit exists, we say the integral is convergent, meaning the area is finite (for positive functions). If the limit doesn't exist, the integral is divergent.

💡 How quickly a function approaches zero matters! For example, $\int_1^{\infty} \frac{1}{x} dx$ diverges because it doesn't approach zero fast enough, but $\int_1^{\infty} \frac{1}{x^2} dx$ converges because it gets smaller quickly.

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Vertical Asymptotes and Type II Integrals

Type II improper integrals handle vertical asymptotes:

If function has a vertical asymptote at point $a$ : $\int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx$
If function has a vertical asymptote at point $b$ : $\int_a^b f(x) dx = \lim_{t \to b^-} \int_a^t f(x) dx$

For discontinuities within an interval, we break the integral at the problem point and evaluate each piece separately: $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$

🔑 Remember these key examples: $\int_0^1 \frac{1}{x^p} dx$ and $\int_1^{\infty} \frac{1}{x^p} dx$ both converge when $p < 1$ and diverge when $p \geq 1$ . These "p-integrals" appear frequently on tests!

When evaluating these integrals, you'll need to compute the limits to justify whether they converge or diverge.

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Access to all documents

Improve your grades

Join milions of students

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Evaluating Complex Improper Integrals

For general improper integrals, you'll need to break them down into simpler Type I and Type II pieces. This divide-and-conquer approach helps tackle complex problems systematically.

The Comparison Test is a powerful tool for determining convergence without calculating the exact value:

If $0 \leq g(x) \leq f(x)$ and $\int g(x)dx$ diverges → $\int f(x)dx$ also diverges
If $0 \leq g(x) \leq f(x)$ and $\int f(x)dx$ converges → $\int g(x)dx$ also converges

🧠 When facing complicated integrals, try bounding them with simpler functions whose convergence you already know. This strategy often saves you from difficult integrations!