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quarto/integrals/improper_integrals.qmd
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@@ -8,8 +8,7 @@ This section uses these add-on packages:
```{julia}
using CalculusWithJulia
using Plots
plotly()
using Plots; plotly()
using SymPy
using QuadGK
```
@@ -21,9 +20,9 @@ using QuadGK
A function $f(x)$ is Riemann integrable over an interval $[a,b]$ if some limit involving Riemann sums exists. This limit will fail to exist if $f(x) = \infty$ in $[a,b]$. As well, the Riemann sum idea is undefined if either $a$ or $b$ (or both) are infinite, so the limit won't exist in this case.
To define integrals with either functions having singularities or infinite domains, the idea of an improper integral is introduced with definitions to handle the two cases above.
To define integrals when a function has a singularity or there is an infinite domains, the idea of an improper integral is introduced.
::: {#fig-area-under-sqrt-x-animation}
```{julia}
#| hold: true
#| echo: false
@@ -39,7 +38,7 @@ function make_sqrt_x_graph(n)
f(x) = 1/sqrt(x)
val = N(integrate(f(x), (x, 1/2^n, b)))
title = L"area under $f$ over $[2^{-%$n}, %$b]$ is $%$(rpad(round(val, digits=2), 4))$"
title = L"area under $f$ over $[2^{-%$n}, %$b]$ is $%$(rpad(round(val, digits=2), 4, '0'))$"
plt = plot(f, range(a, stop=b, length=1000);
@@ -53,11 +52,7 @@ function make_sqrt_x_graph(n)
end
caption = L"""
Area under $1/\sqrt{x}$ over $[a,b]$ increases as $a$ gets closer to $0$. Will it grow unbounded or have a limit?
"""
caption = ""
n = 10
anim = @animate for i=1:n
make_sqrt_x_graph(i)
@@ -69,13 +64,17 @@ plotly()
ImageFile(imgfile, caption)
```
Area under $1/\sqrt{x}$ over $[a,1]$ increases as $a$ gets closer to $0$. Will the area grow unbounded or have a limit?
:::
## Infinite domains
Let $f(x)$ be a reasonable function, so reasonable that for any $a < b$ the function is Riemann integrable, meaning $\int_a^b f(x)dx$ exists.
What needs to be the case so that we can discuss the integral over the entire real number line?
What needs to be the case so that we can discuss the integral of $f(x)$ over the entire real number line?
Clearly something. The function $f(x) = 1$ is reasonable by the idea above. Clearly the integral over $[a,b]$ is just $b-a$, but the limit over an unbounded domain would be $\infty$. Even though limits of infinity can be of interest in some cases, not so here. What will ensure that the area is finite over an infinite region?
@@ -84,26 +83,32 @@ Clearly something. The function $f(x) = 1$ is reasonable by the idea above. Clea
Or is that even the right question. Now consider $f(x) = \sin(\pi x)$. Over every interval of the type $[-2n, 2n]$ the area is $0$, and over any interval, $[a,b]$ the area never gets bigger than $2$. But still this function does not have a well defined area on an infinite domain.
The right question involves a limit. Fix a finite $a$. We define the definite integral over $[a,\infty)$ to be
The right approach involves a limit.
::: {.definition title="Definite integral over an unbounded domain"}
Fix a finite $a$. We define the definite integral over $[a,\infty)$ to be
$$
\int_a^\infty f(x) dx = \lim_{M \rightarrow \infty} \int_a^M f(x) dx,
$$
when the limit exists. Similarly, we define the definite integral over $(-\infty, a]$ through
when the limit exists.
Similarly, we define the definite integral over $(-\infty, a]$ through
$$
\int_{-\infty}^a f(x) dx = \lim_{M \rightarrow -\infty} \int_M^a f(x) dx.
$$
For the interval $(-\infty, \infty)$ we have need *both* these limits to exist, and then:
For the interval $(-\infty, \infty)$ we need *both* these limits to exist, and then:
$$
\int_{-\infty}^\infty f(x) dx = \lim_{M \rightarrow -\infty} \int_M^a f(x) dx + \lim_{M \rightarrow \infty} \int_a^M f(x) dx.
$$
:::
:::{.callout-note}
## Note
@@ -118,7 +123,7 @@ When the integral exists, it is said to *converge*. If it doesn't exist, it is s
$$
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^2}dx = \lim_{M \rightarrow \infty} -\frac{1}{x}\big|_1^M
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^2}dx = \lim_{M \rightarrow \infty} -\frac{1}{x}\Big|_1^M
= \lim_{M \rightarrow \infty} 1 - \frac{1}{M} = 1.
$$
@@ -126,7 +131,7 @@ $$
$$
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^{1/2}}dx = \lim_{M \rightarrow \infty} \frac{x^{1/2}}{1/2}\big|_1^M
\lim_{M \rightarrow \infty} \int_1^M \frac{1}{x^{1/2}}dx = \lim_{M \rightarrow \infty} \frac{x^{1/2}}{1/2}\Big|_1^M
= \lim_{M \rightarrow \infty} 2\sqrt{M} - 2 = \infty.
$$
@@ -152,28 +157,37 @@ $$
for any finite $a$. This is because, $F(M) = e^x$ and this has a limit as $x$ goes to $-\infty$, but not $\infty$.
* Let $f(x) = x e^{-x^2}$. This function has an integral over $[0, \infty)$ and more generally $(-\infty, \infty)$. To see, we note that as it is an odd function, the area from $0$ to $M$ is the opposite sign of that from $-M$ to $0$. So $\lim_{M \rightarrow \infty} (F(M) - F(0)) = \lim_{M \rightarrow -\infty} (F(0) - (-F(\lvert M\lvert)))$. We only then need to investigate the one limit. But we can see by substitution with $u=x^2$, that an antiderivative is $F(x) = (-1/2) \cdot e^{-x^2}$. Clearly, $\lim_{M \rightarrow \infty}F(M) = 0$, so the answer is well defined, and the area from $0$ to $\infty$ is just $1/2$. From $-\infty$ to $0$ it is $-1/2$ and the total area is $0$, as the two sides "cancel" out.
* Let $f(x) = \sin(x)$. Even though $\lim_{M \rightarrow \infty} (F(M) - F(-M) ) = 0$, this function is not integrable. The fact is we need *both* the limit $F(M)$ and $F(-M)$ to exist as $M$ goes to $\infty$. In this case, even though the area cancels if $\infty$ is approached at the same rate, this isn't sufficient to guarantee the two limits exists independently.
* Let $f(x) = x e^{-x^2}$. This function has an integral over $[0, \infty)$ and more generally $(-\infty, \infty)$. To see, we note that as it is an odd function, the area from $0$ to $M$ is the opposite sign of that from $-M$ to $0$. So:
$$
\lim_{M \rightarrow \infty} (F(M) - F(0)) = \lim_{M \rightarrow -\infty} (F(0) - (-F(\lvert M\lvert))).
$$
We only then need to investigate the one limit. But we can see by substitution with $u=x^2$, that an antiderivative is $F(x) = (-1/2) \cdot e^{-x^2}$. Clearly, $\lim_{M \rightarrow \infty}F(M) = 0$, so the answer is well defined, and the area from $0$ to $\infty$ is just $1/2$. From $-\infty$ to $0$ it is $-1/2$ and the total area is $0$, as the two sides "cancel" out.
* Let $f(x) = \sin(x)$. Even though $\lim_{M \rightarrow \infty} (F(M) - F(-M) ) = 0$, this function is not integrable. The fact is we need *both* the limit $F(M)$ and $F(-M)$ to exist as $M$ goes to $\infty$. In this case, even though the area cancels if $\infty$ is approached at the same rate, this isn't sufficient to guarantee the two limits exists independently.
* Will the function $f(x) = 1/(x\cdot(\log(x))^2)$ have an integral over $[e, \infty)$?
* Will the function $f(x) = 1/(x\cdot(\log(x))^2)$ have an integral over $[e, \infty)$?
We first find an antiderivative using the $u$-substitution $u(x) = \log(x)$:
$$
\begin{align*}
\int_e^M \frac{1}{x \log(x)^{2}} dx
= \int_{\log(e)}^{\log(M)} \frac{1}{u^{2}} du
= \frac{-1}{u} \big|_{1}^{\log(M)}
= \frac{-1}{\log(M)} - \frac{-1}{1}
= 1 - \frac{1}{\log(M)}.
&= \int_{\log(e)}^{\log(M)} \frac{1}{u^{2}} du\\
&= \frac{-1}{u} \Big|_{1}^{\log(M)}\\
&= \frac{-1}{\log(M)} - \frac{-1}{1}\\
&= 1 - \frac{1}{\log(M)}.
\end{align*}
$$
As $M$ goes to $\infty$, this will converge to $1$.
* The sinc function $f(x) = \sin(\pi x)/(\pi x)$ does not have a nice antiderivative. Seeing if the limit exists is a bit of a problem. However, this function is important enough that there is a built-in function, `Si`, that computes $\int_0^x \sin(u)/u\cdot du$. This function can be used through `sympy.Si(...)`:
* The sinc function $f(x) = \sin(\pi x)/(\pi x)$ does not have a nice antiderivative. Seeing if the limit exists is a bit of a problem^[Well, we could break the answer into an alternating series with clearly shrinking terms, so this is a bit of an exaggeration.]. However, this function is important enough that there is a built-in function, `Si`, that computes $\int_0^x \sin(u)/u\cdot du$. This function can be used through `sympy.Si(...)`:
```{julia}
@@ -192,13 +206,13 @@ $$
we introduce a trick and rely on some theorems that have not been discussed.
First, we notice that $\Si(x)$ is the value of $I(\alpha)$ when $\alpha=0$ where
First, we notice that $\text{Si}(x)$ is the value of $I(\alpha)$ when $\alpha=0$ where
$$
I(\alpha) = \int_0^\infty \exp(-\alpha t) \frac{\sin(t)}{t} dt
$$
We differentiate $I$ in $\alpha$ to get:
We differentiate $I$ in $\alpha$ to get:^[This is a bit of a fast one, as we move the derivative *inside* the integral.]
$$
\begin{align*}
@@ -217,8 +231,8 @@ $$
&=\sin(t) \frac{-\exp(-\alpha t)}{\alpha} \Big|_0^\infty -
\int_0^\infty \frac{-\exp(-\alpha t)}{\alpha} \cos(t) dt \\
&= 0 + \frac{1}{\alpha} \cdot \int_0^\infty \exp(-\alpha t) \cos(t) dt \\
&= \frac{1}{\alpha} \cdot \cos(t)\frac{-\exp(-\alpha t)}{\alpha} \Big|_0^\infty -
\frac{1}{\alpha} \cdot \int_0^\infty \frac{-\exp(-\alpha t)}{\alpha} (-\sin(t)) dt \\
&= \frac{1}{\alpha} \cdot \cos(t)\frac{-\exp(-\alpha t)}{\alpha} \Big|_0^\infty -\\
&\quad\frac{1}{\alpha} \cdot \int_0^\infty \frac{-\exp(-\alpha t)}{\alpha} (-\sin(t)) dt \\
&= \frac{1}{\alpha^2} - \frac{1}{\alpha^2} \cdot \int_0^\infty \exp(-\alpha t) \sin(t) dt
\end{align*}
$$
@@ -232,7 +246,7 @@ $$
Solving gives the desired integral as
$$
I'(\alpha) = -\frac{1}{\alpha^2} / (1 + \frac{1}{\alpha^2}) = -\frac{1}{1 + \alpha^2}.
I'(\alpha) = -\frac{1}{\alpha^2} / \left(1 + \frac{1}{\alpha^2}\right) = -\frac{1}{1 + \alpha^2}.
$$
@@ -242,14 +256,15 @@ As our question is answered by $I(0)$, we get $I(0) = \tan^{-1}(0) + C = C = \pi
The above argument requires two places where a *limit* is passed inside the integral. The first involved the derivative. The [Leibniz integral rule](https://en.wikipedia.org/wiki/Leibniz_integral_rule) can be used to verify the first use is valid:
:::{.callout-note icon=false}
## Leibniz integral rule
:::{.theorem title="Leibniz integral rule"}
If $f(x,t)$ and the derivative in $x$ for a fixed $t$ is continuous (to be discussed later) in a region containing $a(x) \leq t \leq b(x)$ and $x_0 < x < x_1$ and both $a(x)$ and $b(x)$ are continuously differentiable, then
$$
\frac{d}{dx}\int_{a(x)}^{b(x)} f(x, t) dt =
\int_{a(x)}^{b(x)} \frac{d}{dx}f(x,t) dt +
f(x, b(x)) \frac{d}{dx}b(x) - f(x, a(x)) \frac{d}{dx}a(x).
\begin{align*}
\frac{d}{dx}\int_{a(x)}^{b(x)} f(x, t) dt
&= \int_{a(x)}^{b(x)} \frac{d}{dx}f(x,t) dt \\
&\quad + f(x, b(x)) \frac{d}{dx}b(x) - f(x, a(x)) \frac{d}{dx}a(x).
\end{align*}
$$
:::
@@ -296,7 +311,7 @@ Suppose $a < c$, we define $\int_a^c f(x) dx = \lim_{M \rightarrow c-} \int_a^M
$$
\lim_{M \rightarrow 0+} \int_M^1 \frac{1}{\sqrt{x}} dx
= \lim_{M \rightarrow 0+} \frac{\sqrt{x}}{1/2} \big|_M^1
= \lim_{M \rightarrow 0+} \frac{\sqrt{x}}{1/2} \Big|_M^1
= \lim_{M \rightarrow 0+} 2(1) - 2\sqrt{M} = 2.
$$
@@ -311,7 +326,7 @@ The cases $f(x) = x^{-n}$ for $n > 0$ are tricky to keep straight. For $n > 1$,
$$
\lim_{M \rightarrow 0+} \int_M^1 \frac{1}{x} dx
= \lim_{M \rightarrow 0+} \log(x) \big|_M^1
= \lim_{M \rightarrow 0+} \log(x) \Big|_M^1
= \lim_{M \rightarrow 0+} \log(1) - \log(M) = \infty.
$$
@@ -363,7 +378,7 @@ A probability density is a function $f(x) \geq 0$ which is integrable on $(-\inf
Probability densities are good example of using improper integrals.
* Show that $f(x) = (1/\pi) (1/(1 + x^2))$ is a probability density function.
* Show that $f(x) = (1/\pi) (1/(1 + x^2))$ is a probability density function.
We need to show that the integral exists and is $1$. For this, we use the fact that $(1/\pi) \cdot \tan^{-1}(x)$ is an antiderivative. Then we have:
@@ -376,7 +391,7 @@ $$
and as $\tan^{-1}(x)$ is odd, we must have $F(-\infty) = \lim_{M \rightarrow -\infty} f(M) = -(1/\pi) \cdot \pi/2$. All told, $F(\infty) - F(-\infty) = 1/2 - (-1/2) = 1$.
* Show that $f(x) = 1/(b-a)$ for $a \leq x \leq b$ and $0$ otherwise is a probability density.
* Show that $f(x) = 1/(b-a)$ for $a \leq x \leq b$ and $0$ otherwise is a probability density.
The integral for $-\infty$ to $a$ of $f(x)$ is just an integral of the constant $0$, so will be $0$. (This is the only constant with finite area over an infinite domain.) Similarly, the integral from $b$ to $\infty$ will be $0$. This means:
@@ -389,10 +404,10 @@ $$
(One might also comment that $f$ is Riemann integrable on any $[0,M]$ despite being discontinuous at $a$ and $b$.)
* Show that if $f(x)$ is a probability density then so is $f(x-c)$ for any $c$.
* Show that if $f(x)$ is a probability density then so is $f(x-c)$ for any $c$.
We have by the $u$-substitution
We have by the $u$-substitution $u(x)=x-c$ that
$$
@@ -402,7 +417,7 @@ $$
The key is that we can use the regular $u$-substitution formula provided $\lim_{M \rightarrow \infty} u(M) = u(\infty)$ is defined. (The *informal* notation $u(\infty)$ is defined by that limit.)
* If $f(x)$ is a probability density, then so is $(1/h) f((x-c)/h)$ for any $c, h > 0$.
* If $f(x)$ is a probability density, then so is $(1/h) f((x-c)/h)$ for any $c, h > 0$.
Again, by a $u$ substitution with, now, $u(x) = (x-c)/h$, we have $du = (1/h) \cdot dx$ and the result follows just as before:
@@ -412,10 +427,10 @@ $$
\int_{-\infty}^\infty \frac{1}{h}f(\frac{x-c}{h})dx = \int_{u(-\infty)}^{u(\infty)} f(u) du = \int_{-\infty}^\infty f(u) du = 1.
$$
* If $F(x) = 1 - e^{-x}$, for $x \geq 0$, and $0$ otherwise, find $f(x)$.
* If $F(x) = 1 - e^{-x}$, for $x \geq 0$, and $0$ otherwise, find $f(x)$.
We want to just say $F'(x)= e^{-x}$ so $f(x) = e^{-x}$. But some care is needed. First, that isn't right. The derivative for $x<0$ of $F(x)$ is $0$, so $f(x) = 0$ if $x < 0$. What about for $x>0$? The derivative is $e^{-x}$, but is that the right answer? $F(x) = \int_{-\infty}^x f(u) du$, so we have to at least discuss if the $-\infty$ affects things. In this case, and in general the answer is *no*. For any $x$ we can find $M < x$ so that we have $F(x) = \int_{-\infty}^M f(u) du + \int_M^x f(u) du$. The first part is a constant, so will have derivative $0$, the second will have derivative $f(x)$, if the derivative exists (and it will exist at $x$ if the derivative is continuous in a neighborhood of $x$).
We want to just say $F'(x)= e^{-x}$ so $f(x) = e^{-x}$. But some care is needed. First, that isn't right. The derivative for $x<0$ of $F(x)$ is $0$, so $f(x) = 0$ if $x < 0$. What about for $x>0$? The derivative is $e^{-x}$, but is that the right answer? $F(x) = \int_{-\infty}^x f(u) du$, so we have to at least discuss if the $-\infty$ affects things. In this case, and in general, the answer is *no*. For any $x$ we can find $M < x$ so that we have $F(x) = \int_{-\infty}^M f(u) du + \int_M^x f(u) du$. The first part is a constant, so will have derivative $0$, the second will have derivative $f(x)$, if the derivative exists (and it will exist at $x$ if the derivative is continuous in a neighborhood of $x$).
Finally, at $x=0$ we have an issue, as $F'(0)$ does not exist. The left limit of the secant line approximation is $0$, the right limit of the secant line approximation is $1$. So, we can take $f(x) = e^{-x}$ for $x > 0$ and $0$ otherwise, noting that redefining $f(x)$ at a point will not effect the integral as long as the point is finite.
@@ -424,16 +439,45 @@ Finally, at $x=0$ we have an issue, as $F'(0)$ does not exist. The left limit of
## Application to series
In this application, we compare a series to a related integral to decide convergence or divergence of the series.
In this application, we compare a series to a related integral to decide convergence or divergence of the series. @fig-integral-test-figure motivates the following theorem.
:::{.theorem title="The integral test"}
:::{.callout-note appearance="minimal"}
#### The integral test
Consider a continuous, monotone decreasing function $f(x)$ defined on some interval of the form $[N,\infty)$. Let $a_n = f(n)$ and $s_n = \sum_{k=N}^n a_n$.
* If $\int_N^\infty f(x) dx < \infty$ then the partial sums converge.
* If $\int_N^\infty f(x) dx = \infty$ then the partial sums diverge.
:::
::: {#fig-integral-test-figure}
```{julia}
#| echo: false
let
# integral test
gr()
f(x) = 1/x
p1 = plot(; legend=false, framestyle=:origin, xticks=1:8, yaxis=([], false))
p2 = plot(; legend=false, framestyle=:origin, xticks=1:8, yaxis=([], false))
plot!(p1, f, 0.75, 8.25; line=(1, :black))
plot!(p2, f, 0.75, 8.25; line=(1, :black))
for k in 1:7
plot!(p1, [(k,0), (k+1,0), (k+1, f(k)), (k, f(k)), (k,0)]; line=(1, :black, :dot))
annotate!(p1, [(k+1/2, f(k+1)/2, latexstring("a_{$k}"))])
end
for k in 1:7
plot!(p2, [(k,0), (k+1,0), (k+1, f(k+1)), (k, f(k+1)),(k,0)]; line=(1, :black, :dot))
annotate!(p2, [(k+1/2, f(k+2)/2, latexstring("a_{$(k+1)}"))])
end
plotly()
plot(p1, p2)
end
```
Illustration of the integral test, where a series $a_1 + a_2 + \cdots$ is bounded *below* by $\int_1^\infty f(x)dx$ and the series $a_2 + a_3 + \cdots$ is bounded *above* by $\int_1^\infty f(x)dx$ where $a_i=f(i)$. In either case, converge/divergence of the integral forces convergence/divergence of the series.
:::
By the monotone nature of $f(x)$, we have on any interval of the type $[i, i+1)$ for $i$ an integer, that $f(i) \geq f(x) \geq f(i+1)$ when $x$ is in the interval. For integrals, this leads to
$$
@@ -491,8 +535,7 @@ That this is finite shows the series converges.
The integral of a power series can be computed easily for some $x$:
:::{.callout-note appearance="minimal"}
### The integral of a power series
:::{.theorem title="The integral of a power series"}
Suppose $f(x) = \sum_n a_n (x-c)^n$ is a power series about $x=c$ with radius of convergence $r > 0$. [Then](https://en.wikipedia.org/wiki/Power_series#Differentiation_and_integration) the limits of the integral and the sum can be switched around when $x$ is within the radius of convergence:
@@ -501,7 +544,7 @@ $$
\int f(x) dx
&= \int \sum_n a_n(x-c)^n dx\\
&= \sum_{n=0}^\infty \int a_n(x-c)^n dx\\
= \sum_{n=0}^\infty a_n \frac{(x-c)^{n+1}}{n+1}
&= \sum_{n=0}^\infty a_n \frac{(x-c)^{n+1}}{n+1}
\end{align*}
$$
@@ -647,14 +690,39 @@ val, _ = quadgk(f , 0, 1, 2)
numericq(val)
```
###### Question
Consider the integral $\int_{-\infty}^\infty f(x) dx$. We do a change of variable with $x = t/(1-t^2)$. This gives:
$$
\int_a^b f\left(\frac{t}{1-t^2}\right) \frac{t^2 + 1}{t^2 - 1} dt
$$
What are the values of $a$ and $b$?
```{julia}
#| echo: false
choices = [
L"$a=-1$ and $b=1$",
L"$a=1$ and $b=-1$",
L"$a=1$ and $b=0$",
L"$a=0$ and $b=1$",
]
answer = 1
explanation = "As ``t`` goes to ``1`` from the left, ``x`` goes to what?"
buttonq(choices, answer; explanation)
```
###### Question
From the relationship that if $0 \leq f(x) \leq g(x)$ then $\int_a^b f(x) dx \leq \int_a^b g(x) dx$ it can be deduced that
* if $\int_a^\infty f(x) dx$ diverges, then so does $\int_a^\infty g(x) dx$.
* if $\int_a^\infty g(x) dx$ converges, then so does $\int_a^\infty f(x) dx$.
* if $\int_a^\infty f(x) dx$ diverges, then so does $\int_a^\infty g(x) dx$.
* if $\int_a^\infty g(x) dx$ converges, then so does $\int_a^\infty f(x) dx$.
Let $f(x) = \lvert \sin(x)/x^2 \rvert$.
@@ -670,8 +738,8 @@ choices =[
"It is convergent",
"It is divergent",
"Can't say"]
answ = 1
radioq(choices, answ, keep_order=true)
answer = 1
radioq(choices, answer, keep_order=true)
```
---
@@ -690,8 +758,8 @@ choices =[
"It is convergent",
"It is divergent",
"Can't say"]
answ = 3
radioq(choices, answ, keep_order=true)
answer = 3
radioq(choices, answer, keep_order=true)
```
---
@@ -707,8 +775,8 @@ choices =[
"It is convergent",
"It is divergent",
"Can't say"]
answ = 2
radioq(choices, answ, keep_order=true)
answer = 2
radioq(choices, answer, keep_order=true)
```
---
@@ -724,8 +792,8 @@ choices =[
"It is convergent",
"It is divergent",
"Can't say"]
answ = 1
radioq(choices, answ, keep_order=true)
answer = 1
radioq(choices, answer, keep_order=true)
```
---
@@ -741,8 +809,8 @@ choices =[
"It is convergent",
"It is divergent",
"Can't say"]
answ = 1
radioq(choices, answ, keep_order=true)
answer = 1
radioq(choices, answer, keep_order=true)
```
###### Question
@@ -759,8 +827,8 @@ choices = [
"``\\int_0^1 u^{2/3} \\cdot du``",
"``\\int_0^\\infty 1/u \\cdot du``"
]
answ = 1
radioq(choices, answ)
answer = 1
radioq(choices, answer)
```
###### Question