Wize AP Calculus (BC) Textbook > Improper Integrals

Comparison Theorem for Improper Integrals

Popular Courses

AP Exam Prep

0:00 / 0:00

Comparison Theorem for Improper Integrals
Common Improper Integrals 
1. ∫a∞1xp  ⁣dx{\displaystyle\int^\infty_a}\frac{1}{x^p}\de{x}∫a∞​xp1​ dx
If p≤1p\le1p≤1: diverges
If p>1p>1p>1: converges

2. ∫0a1xp  ⁣dx{\displaystyle\int^a_0}\frac{1}{x^p}\de{x}∫0a​xp1​ dx
If p<1p<1p<1: converges
If p≥1p\ge1p≥1: diverges
Comparison Theorem
Suppose that fff and ggg are continuous and f(x)≥g(x)≥0f(x)\ge g(x)\ge 0f(x)≥g(x)≥0 for x≥a.x \ge a.x≥a.
a. If ∫a∞f(x)  ⁣dx{\displaystyle\int^\infty_a}f(x)\de{x}∫a∞​f(x) dx converges, then ∫a∞g(x)  ⁣dx{\displaystyle\int^\infty_a}g(x)\de{x}∫a∞​g(x) dx converges.
b. If ∫a∞g(x)  ⁣dx{\displaystyle\int^\infty_a}g(x)\de{x}∫a∞​g(x) dx diverges, then ∫a∞f(x)  ⁣dx{\displaystyle\int^\infty_a}f(x)\de{x}∫a∞​f(x) dx diverges. 

Notes:
The theorem only tells you if an improper integral diverges or converges, not the value it converges to
Often, we will use the common improper integrals and the Comparison Theorem

Practice: Comparison Test
Determine whether the following improper integral is convergent or divergent: 
∫1∞xx2+x  ⁣dx{\displaystyle\int}^\infty_1\frac{\sqrt{x}}{x^2+x}\de{x}∫1∞​x2+xx​​ dx
  

It converges

It diverges

I don't know

Practice: Comparison Theorem
Determine whether ∫02arctan⁡x5x dx{\displaystyle \int^2_0}\frac{\arctan x^5}{\sqrt{x}}\ dx∫02​x​arctanx5​ dx converges or diverges.

It converges

It diverges

I don't know

Practice: Comparison Theorem
Determine if ∫01/41x2+x5  ⁣dx{\displaystyle\int^{1/4}_0}\dfrac{1}{x^2+x^5}\de{x}∫01/4​x2+x51​ dx  is convergent or divergent. 
  

It converges

It diverges

I don't know