Calculus BC · Unit 6
Turn an infinite limit or a blow-up into an ordinary limit, then decide whether that limit is finite.
Replace the infinite limit by a letter
\(\int_a^{\infty} f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx\)
The integral converges when that limit is a finite number.
A bad point inside the interval needs the same treatment
If \(f\) blows up at an endpoint, move a letter in toward that endpoint.
Power tails
\(\int_1^{\infty}\dfrac{1}{x^p}\,dx\) converges when \(p>1\) and diverges when \(p\le 1\).
On \((0,1]\), \(\int_0^1 x^{-p}\,dx\) converges when \(p<1\).
A comparison can settle the question
If \(0\le f(x)\le g(x)\) and the integral of \(g\) converges, then the integral of \(f\) converges.
One infinite limit and a power you can antidifferentiate.
\(\lim_{b\to\infty}\int_1^b x^{-2}\,dx\)
\(\lim_{b\to\infty}\left[-\dfrac{1}{x}\right]_1^b\)
\(\lim_{b\to\infty}\left(-\dfrac{1}{b}+1\right)\)
\(1\)
The limit is finite, so the integral converges.
Answer: \(1\)
\(\lim_{b\to\infty}\left[\ln x\right]_1^b\)
\(\lim_{b\to\infty}\ln b\)
The limit is infinite.
The integral diverges.
Answer: diverges
\(\lim_{b\to\infty}\int_0^b e^{-3x}\,dx\)
\(\lim_{b\to\infty}\left[-\dfrac{1}{3}e^{-3x}\right]_0^b\)
\(\lim_{b\to\infty}\left(-\dfrac{1}{3}e^{-3b}+\dfrac{1}{3}\right)\)
\(\dfrac{1}{3}\)
Answer: \(\dfrac{1}{3}\)
The integrand is unbounded at \(x=0\).
\(\lim_{a\to 0^+}\int_a^1 x^{-1/2}\,dx\)
\(\lim_{a\to 0^+}\left[2x^{1/2}\right]_a^1\)
\(\lim_{a\to 0^+}(2-2\sqrt{a})\)
\(2\)
Answer: \(2\)
\(\lim_{a\to 0^+}\left[\ln x\right]_a^1\)
\(\lim_{a\to 0^+}(-\ln a)\)
The limit is infinite.
The integral diverges.
Answer: diverges
\(1/\sqrt{x}\) has \(p=1/2\le 1\) on \([1,\infty)\), so it diverges.
\(1/x^{3/2}\) has \(p=3/2>1\), so it converges.
A constant tail and \(1/x^2\) on \((0,1]\) both diverge.
Answer: B. \(\int_1^{\infty} x^{-3/2}\,dx\)
Move the bad endpoint, or compare with a known tail.
\(u=x-1\)
When \(x=2\), \(u=1\). The upper limit still goes to infinity.
\(\lim_{b\to\infty}\left[-\dfrac{1}{u}\right]_1^{b-1}\)
\(\lim_{b\to\infty}\left(-\dfrac{1}{b-1}+1\right)\)
\(1\)
Answer: \(1\)
\(\lim_{a\to-\infty}\left[\dfrac{1}{2}e^{2x}\right]_a^0\)
\(\dfrac{1}{2}-\lim_{a\to-\infty}\dfrac{1}{2}e^{2a}\)
The exponential goes to 0.
\(\dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
The integrand blows up at \(x=1\).
\(u=1-x\) and \(du=-dx\).
The limit \(x\to 1^-\) becomes \(u\to 0^+\), and \(x=0\) becomes \(u=1\).
\(\lim_{c\to 0^+}\int_c^1 u^{-1/2}\,du\)
\(\lim_{c\to 0^+}\left[2u^{1/2}\right]_c^1\)
\(2\)
Answer: \(2\)
For \(x\ge 1\), \(x^2+4>x^2\).
\(0<\dfrac{1}{x^2+4}<\dfrac{1}{x^2}\)
\(\int_1^{\infty} x^{-2}\,dx\) converges.
A smaller positive function converges too.
Answer: converges
\(\int \dfrac{1}{x^2+4}\,dx=\dfrac{1}{2}\arctan\dfrac{x}{2}\)
\(\lim_{b\to\infty}\dfrac{1}{2}\left(\arctan\dfrac{b}{2}-\arctan\dfrac{1}{2}\right)\)
\(\dfrac{1}{2}\left(\dfrac{\pi}{2}-\arctan\dfrac{1}{2}\right)\)
Answer: \(\dfrac{1}{2}\left(\dfrac{\pi}{2}-\arctan\dfrac{1}{2}\right)\)
\(\lim_{b\to\infty}\left[-e^{-x}\right]_0^b\)
\(\lim_{b\to\infty}(-e^{-b}+1)\)
\(1\)
Answer: \(1\)
Two bad ends, or a comparison that points the right way.
The integrand is unbounded at 0, and the interval is also infinite.
Split: \(\int_0^1 + \int_1^{\infty}\).
Near 0, \(1+x\) is about 1, so the piece behaves like \(x^{-1/2}\), which converges on \((0,1]\).
For large \(x\), the integrand behaves like \(x^{-1/2}/x=x^{-3/2}\), and \(p=3/2>1\), so that tail converges.
Both pieces converge, so the original integral converges.
Answer: converges
\(u=x+2\)
\(\lim_{b\to\infty}\int_2^{b+2} u^{-3}\,du\)
\(\lim_{b\to\infty}\left[-\dfrac{1}{2u^2}\right]_2^{b+2}\)
\(0-\left(-\dfrac{1}{2\cdot 4}\right)\)
\(\dfrac{1}{8}\)
Answer: \(\dfrac{1}{8}\)
\(u=\ln x\)
\(du=\dfrac{1}{x}\,dx\)
When \(x=2\), \(u=\ln 2\). The upper limit goes to infinity.
\(\lim_{b\to\infty}\left[\ln u\right]_{\ln 2}^{\ln b}\)
This is \(\lim(\ln(\ln b)-\ln(\ln 2))\), which is infinite.
The integral diverges.
Answer: diverges
The comparison test needs the larger function’s integral to converge.
\(\int_1^{\infty} x^{-1/2}\,dx\) diverges.
Being smaller than a divergent integral gives no conclusion.
The harmonic integral diverges by its own limit.
Answer: the comparison is in the wrong direction
\(p=2/3<1\), so a finite value is expected.
\(\lim_{a\to 0^+}\left[\dfrac{x^{1/3}}{1/3}\right]_a^1\)
\(\lim_{a\to 0^+} 3\left(1-a^{1/3}\right)\)
\(3\)
Answer: \(3\)
\(1/x^2\) is unbounded at \(x=0\), which is inside \([-1,1]\).
The antiderivative rule applies only on an interval where the function is continuous.
Split at 0. Each piece \(\int x^{-2}\,dx\) diverges.
The correct conclusion is that the improper integral diverges.
Answer: diverges
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