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An integrand that blows up on a bounded interval makes the integral improper too, and near zero the test reverses: $\int_0^{1}x^{-p}dx$ converges exactly when $p < 1$. A singularity inside the interval has to be split at.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can recognise an integral that is improper because its integrand is unbounded, replace the bad endpoint by a one-sided limit, and evaluate it. You can apply the $p$-test near zero and say why it is the reverse of the test at infinity, give the value of a convergent one, and find a singularity sitting inside an interval and split the integral at it rather than integrating through.
You can evaluate an integral that runs to infinity by writing it as a limit, and you know the $p$-test that settles it. This lesson is the same technique applied to the other thing that can go wrong — and the same boundary exponent, with the two sides of it exchanged.
Singular endpoint: a limit of integration where the integrand is unbounded.
One-sided limit: the limit taken from inside the interval, written $t \to 0^{+}$ — the only side on which the integral was ever defined.
Splitting at a singularity: breaking an interval at an interior bad point so that each half has its trouble at an end.
The $p$-test near zero: $\int_0^1 x^{-p}dx$ converges exactly when $p < 1$.
$\int_0^1 \dfrac{dx}{\sqrt{x}}$ has two perfectly ordinary limits and is still improper, because the integrand is unbounded at the left end. The cure is the same as before:
$$\int_0^1 f = \lim_{t\to 0^{+}}\int_t^1 f,$$
approaching the bad endpoint from inside the interval.
The test reverses. With the same antiderivative $\dfrac{x^{1-p}}{1-p}$, what matters near zero is whether $1-p$ is positive. It is when $p < 1$, and then $t^{1-p}\to 0$ and the integral converges to $\dfrac{1}{1-p}$. So $\int_0^1 x^{-p}dx$ converges exactly when $p < 1$ — the opposite of the condition at infinity.
That is not a coincidence to be memorised twice. A large exponent makes $x^{-p}$ small far out and enormous near zero; the exponent that tames one end sharpens the other. And at $p = 1$ the antiderivative is $\ln x$, which runs away at both ends, so that single exponent diverges in both directions — the one place the two tests agree.
A singularity in the middle has to be found. $\int_{-1}^{1}\dfrac{dx}{x^2}$ looks ordinary and is not: the integrand is unbounded at $0$. The interval must be split there and each half tested on its own. An antiderivative evaluated straight through gives $-2$ — a negative number for the area under a positive function, and a confident, checkable lie.
Another way: picture
Picture a wall at the left edge of the region rather than a tail stretching to the right. The region is infinitely tall and only one unit wide, and the question is whether it is so thin so quickly that its area is still finite. Sometimes it is. A tall thin spike can have less paint in it than a low wide rectangle.
Another way: steps
| The integral | Converges when | Value then |
|---|---|---|
| $\int_1^{\infty}x^{-p}dx$ | $p > 1$ | $\dfrac{1}{p-1}$ |
| $\int_0^{1}x^{-p}dx$ | $p < 1$ | $\dfrac{1}{1-p}$ |
| either, at $p = 1$ | never | the antiderivative is $\ln x$ |
Read the first two rows together and the symmetry does the remembering for you: the same two numbers, subtracted the other way round, on the other side of the same boundary.
One consequence worth noticing: $\int_0^{\infty}x^{-p}dx$ never converges, for any $p$ at all. Whatever exponent saves one end ruins the other.
$\int_{-1}^{1}\dfrac{dx}{x^{2}}$ with the antiderivative $-\dfrac1x$ evaluated straight through gives $-1 - 1 = -2$. The integrand is positive everywhere it is defined, so an area of $-2$ is impossible, and the true answer is that the integral diverges.
The lesson is not that the arithmetic was careless. The fundamental theorem requires the integrand to be continuous on the closed interval, and here it is not, so the theorem was never available. Every rule in calculus has hypotheses, and a rule applied outside them produces an answer that looks exactly like a real one.
Using the test from infinity near zero. They are opposites. $\int_0^1 \dfrac{dx}{x^{2}}$ diverges, even though $\int_1^{\infty}\dfrac{dx}{x^{2}}$ converges to $1$.
Not noticing the integral is improper at all. Both limits being numbers proves nothing. Check whether the integrand is defined and bounded at every point of the interval, including the ends.
Integrating straight through an interior singularity. The fundamental theorem needs a continuous integrand. Split at the bad point, or the answer is not merely wrong but meaningless.
$\int_0^1 \dfrac{dx}{\sqrt{x}} = \lim_{t\to0^{+}}\int_t^1 x^{-1/2}dx$.
The trouble is at the left end.
$= \lim_{t\to0^{+}}\left[2\sqrt{x}\right]_t^1 = \lim_{t\to0^{+}}(2 - 2\sqrt{t})$.
Ordinary from here.
$= 2$, which is $\dfrac{1}{1-p}$ with $p = \tfrac12$. Infinitely tall, and its area is $2$.
$\int_{-1}^{1}\dfrac{dx}{x^{2}}$: the integrand is unbounded at $0$, strictly inside the interval.
Nothing in the limits says so.
Split it: $\int_{-1}^{0} + \int_{0}^{1}$, and test each half separately.
Now each has its trouble at an end.
Each is $\int_0^1 x^{-2}dx$ in disguise, with $p = 2 > 1$, so each diverges — and so does the whole. Evaluating straight through would have given $-2$.
The trouble is at $0$, and the exponent is $p = \tfrac23$.
Identify the bad end first.
Near zero the test asks whether $p < 1$, and $\tfrac23$ is.
So it converges.
The value is $\dfrac{1}{1-p} = \dfrac{1}{1/3} = 3$ — and note that the same exponent would make the integral to infinity diverge.
For each exponent, say what $\int_1^{\infty}x^{-p}dx$ does and what $\int_0^{1}x^{-p}dx$ does.
| From 1 to infinity | From 0 to 1 | |
|---|---|---|
| $p = \tfrac12$ | ||
| $p = 1$ | ||
| $p = 3$ |
Does $\displaystyle\int_0^{1} \dfrac{dx}{x^{\frac{2}{3}}}$ converge or diverge?
$\displaystyle\int_0^{1} \dfrac{dx}{x^{\frac{2}{3}}}$ converges. What is its value?
Answer:
Put the steps of evaluating $\displaystyle\int_0^{2} \dfrac{dx}{\sqrt{x}}$ into the order you do them.
Number the steps in order (write the number in the box):
$\displaystyle\int_0^{1} x^{-\frac{1}{5}}\,dx = \dfrac{A}{B}$ in lowest terms. Give $A$, then $B$.
A = p, B = q
Select every integral below that is improper.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For each exponent, say what $\int_1^{\infty}x^{-p}dx$ does and what $\int_0^{1}x^{-p}dx$ does.
| From 1 to infinity | From 0 to 1 | |
|---|---|---|
| $p = \tfrac12$ | ||
| $p = 1$ | ||
| $p = 2$ |
You can handle an integral whose integrand is unbounded. Without looking: for which exponents does the integral near zero converge, why is that the reverse of the test at infinity, and what must be done when the bad point is in the middle?
10. Your turn: does $\int_0^{1}\dfrac{dx}{x^{2/3}}$ converge?, step 3