Concept understanding — Improper Integral Evaluation
Improper Integral Evaluation
The Intuition First
You already know how to integrate over a finite interval: ∫13f(x)dx is the area under the curve from x=1 to x=3. But what if the region stretches to infinity, or the function shoots up to infinity somewhere in the interval?
That is what improper integrals handle — two situations that break the ordinary rules:
Infinite limits — integrating up to ∞ or down to −∞.
Infinite discontinuities — the function blows up at some point of the interval.
The core idea is the same in both cases: replace the "bad" point with a limit. Integrate up to a finite value, then let that value approach the trouble spot. If the result approaches a finite number, the integral converges; if it grows without bound, it diverges.
The Precise Definitions
Type 1: Infinite limits
∫a∞f(x)dx=limb→∞∫abf(x)dx
∫−∞bf(x)dx=lima→−∞∫abf(x)dx
For a doubly-infinite integral, split at a convenient point c and require both pieces to converge:
∫−∞∞f(x)dx=∫−∞cf(x)dx+∫c∞f(x)dx
Type 2: Infinite discontinuities
If f has a vertical asymptote at an endpoint, approach it from inside the interval:
If f blows up at x=a: ∫abf(x)dx=t→a+lim∫tbf(x)dx
If f blows up at x=b: ∫abf(x)dx=t→b−lim∫atf(x)dx
If the blow-up is at an interior point c, split at c and treat each side separately.
Never treat an improper integral as an ordinary one. Blindly applying the Fundamental Theorem across a discontinuity gives wrong answers. Always first check: is the integrand defined and finite on the whole interval? …
These are four ordinary definite integrals: power rule for (i), substitution for (ii) and (iv), partial fractions for (iii). The values are 319, 9919, log2732, and 81.
Each part is a proper definite integral (the integrand is finite on the whole interval), so we find an antiderivative and apply F(b)−F(a). The only skill is spotting the right technique for each.
(i) ∫23x2dx
Straight power rule: ∫xndx=n+1xn+1 with n=2.
∫23x2dx=[3x3]23=333−323=327−8=319.
(ii) ∫49(30−x3/2)2xdx
The derivative of the inner expression 30−x3/2 is −23x — a constant multiple of the numerator, which flags a substitution.
Method: Identify-the-Technique for Each Definite Integral, Then Apply Limits
Use this for a set of definite integrals of different types: choose the antiderivative technique per integrand, then evaluate with the Fundamental Theorem, ∫abf=F(b)−F(a).
Steps
Step 1: Classify each integrand.
Match to a technique: a plain power → power rule; a composite with its derivative present → substitution; a proper rational function → partial fractions.
Step 2: For a substitution, change the limits too. …
Mistake 1: Keeping the old x-limits after substituting.
Why it's wrong: once you change to u, the limits must become u-values; using the x-limits gives a wrong number. Correct approach: convert limits with u=g(a), u=g(b).