Type 2: Discontinuous Integrands
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.7: “Improper Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-7-improper-integrals |
| Textbook used in class | Stewart, Calculus, Section 7.8: “Improper Integrals” |
Opening Scenario
The FTC requires the integrand to be continuous on the closed interval $[a, b]$. What if the integrand has a vertical asymptote at one of the endpoints or somewhere in the interior? The integral $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ involves a function that blows up at $x = 0$, yet the region under the curve might still have finite area. These Type 2 improper integrals are handled by the same limit strategy as Type 1, but the limit approaches the point of discontinuity.
Quick Reference
Type 2 -- discontinuity at $x = b$: $$\int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx.$$
Type 2 -- discontinuity at $x = a$: $$\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx.$$
Type 2 -- discontinuity at interior point $x = c$, $a < c < b$: $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx.$$
Both halves must converge separately.
$p$-test near 0 (or any singularity): $$\int_0^1\frac{dx}{x^p} \text{ converges if } p < 1, \text{ diverges if } p \geq 1.$$
Key Concepts
1. Locating the Discontinuity
Before integrating, identify where $f$ is discontinuous on $[a, b]$:
- At an endpoint $a$ or $b$: approach from the interior.
- At an interior point $c$: split the integral and evaluate each half.
Never apply FTC directly when the integrand has a singularity on $[a, b]$; the resulting “answer” may be wrong or meaningless.
2. The Limit Strategy
If $f \to +\infty$ as $x \to a^+$: $$\int_a^b f\,dx = \lim_{t\to a^+}\int_t^b f\,dx = \lim_{t\to a^+}[F(b) - F(t)].$$
The integral converges if $F(t)$ has a finite limit as $t \to a^+$.
3. The $p$-Test Near Zero
$$\int_0^1\frac{dx}{x^p}$$
- $p < 1$: integrand goes to $+\infty$ at $x = 0$, but not fast enough; converges to $\dfrac{1}{1-p}$.
- $p \geq 1$: diverges.
This is the opposite threshold from the Type 1 $p$-test ($\int_1^{\infty}$ converges for $p > 1$).
“A function that blows up at an endpoint always gives a divergent integral.” Not true. $\int_0^1 x^{-1/2}\,dx$ converges to 2 even though $x^{-1/2} \to \infty$ as $x \to 0^+$. The function blows up, but it does so gently enough ($p = 1/2 < 1$) that the area remains finite.
Worked Example
Evaluate $\displaystyle\int_0^1\frac{1}{\sqrt{x}}\,dx$.
The integrand $x^{-1/2}$ has a vertical asymptote at $x = 0$.
$$\int_0^1 x^{-1/2}\,dx = \lim_{t\to 0^+}\int_t^1 x^{-1/2}\,dx = \lim_{t\to 0^+}\left[2\sqrt{x}\right]_t^1 = \lim_{t\to 0^+}(2 - 2\sqrt{t}) = 2.$$
Converges to 2. ($p = 1/2 < 1$, consistent with the $p$-test.)
Evaluate $\displaystyle\int_0^1\frac{dx}{x}$.
$$\int_0^1\frac{dx}{x} = \lim_{t\to 0^+}\int_t^1\frac{dx}{x} = \lim_{t\to 0^+}\ln(x)\Big|_t^1 = \lim_{t\to 0^+}(0 - \ln t) = \infty.$$
Diverges. ($p = 1 \geq 1$, consistent with the $p$-test.)
Evaluate $\displaystyle\int_0^3\frac{dx}{x-1}$ (interior singularity at $x = 1$).
Split at $x = 1$:
$\displaystyle\int_0^1\frac{dx}{x-1} = \lim_{t\to 1^-}\ln|x-1|\Big|_0^t = \lim_{t\to 1^-}(\ln|t-1|-\ln 1) = \lim_{t\to 1^-}\ln|t-1| = -\infty$.
This half diverges, so the full integral diverges.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Applying FTC without checking for discontinuities | Blindly computing $\left[\ln|x-1|\right]_0^3 = \ln 2 - \ln 1 = \ln 2$ | The integrand blows up at $x=1 \in [0,3]$; split the integral and take limits |
| Applying the Type 1 $p$-test to Type 2 integrals | Claiming $\int_0^1 x^{-2}\,dx$ converges since $p=2>1$ | For Type 2 (near 0), convergence requires $p<1$; $p=2$ gives a divergent integral |
| Forgetting to check both halves of an interior split | Computing only $\int_0^c$ and not $\int_c^b$ | Both halves must converge separately; if one diverges, the whole integral diverges |
Leveled Practice
Level 1 -- Endpoint Discontinuity
Problem 1. Evaluate $\displaystyle\int_0^4\frac{dx}{\sqrt{4-x}}$.
Show answer
Singularity at $x = 4$ (the upper limit).
$\displaystyle\lim_{t\to 4^-}\int_0^t\frac{dx}{\sqrt{4-x}} = \lim_{t\to 4^-}\left[-2\sqrt{4-x}\right]_0^t = \lim_{t\to 4^-}(-2\sqrt{4-t}+2\sqrt{4}) = 0 + 4 = 4$.
Converges to 4.
Level 2 -- Interior Discontinuity
Problem 2. Determine whether $\displaystyle\int_{-1}^{2}\frac{1}{x^2}\,dx$ converges. Identify the issue first.
Show answer
$1/x^2$ has a vertical asymptote at $x = 0$, which lies inside $[-1, 2]$.
Split: $\int_{-1}^0\frac{dx}{x^2} = \lim_{t\to 0^-}\left[-\frac{1}{x}\right]_{-1}^t = \lim_{t\to 0^-}(-\frac{1}{t}-1) = +\infty$.
This half diverges, so the integral diverges.
(A common error is computing $\left[-\frac{1}{x}\right]_{-1}^2 = -\frac{1}{2}+1 = \frac{1}{2}$ without noticing the singularity.)
Mastery Checklist
Mental Model
A vertical asymptote in the integrand is like a spike in the terrain. The question is whether the spike is narrow enough that the area underneath it is finite. A spike that approaches infinity as $x^{-p}$ is narrow enough ($p < 1$) or too wide ($p \geq 1$). The faster the function grows (the larger $p$ is), the wider the spike is in area terms -- counterintuitively, because the very fast growth is concentrated near a single point, which seems thin, but the area is determined by how fast $p$ drags the spike upward vs. how thin the region is.
Connections
Looking back
- Type 1 improper integrals (Section 7.8): Same limit strategy; the limit approaches the singularity point instead of $\pm\infty$.
- One-sided limits (Section 2.2): The limits $t \to a^+$ and $t \to b^-$ used in the definition.
Looking ahead
- Comparison test for integrals (Section 7.8): Determines convergence without needing an explicit antiderivative.
Back to Techniques of Integration | Next: Comparison Test for Integrals