Category Archives: Integration

November 20, 2024 · 6:57 am

More Integration

I went down a rabbit hole while reading An Imaginary Tale by Paul J Nahin and I decided I wanted to do this…

$\begin{equation*}\int_0^1{x^x dx}\end{equation}$

$\begin{equation*}x^x=e^{ln(x^x)}\end{equation}$

$\begin{equation*}x^x=e^{xln(x)}\end{equation}$

The power series expansion of $e^x$ is

$\begin{equation*}e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=1+xln(x)+\frac{1}{2!}(xln(x))^2+\frac{1}{3!}(xln(x))^3+...\end{equation}$

$\begin{equation*}\therefore e^{xln(x)}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n)\end{equation}$

Hence $\int_0^1{x^x dx}=\int_0^1(\Sigma_{n=0}^{\infty}(\frac{1}{n!}(xln(x))^n dx)$

$\begin{equation*}=\Sigma_{n=0}^{\infty}(\frac{1}{n!}\int_0^1xln(x))^n dx)\end{equation}$

Let’s consider the integral

(1) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx\end{equation*}$

Let $u=ln(x)$ then $\frac{du}{dx}=\frac{1}{x}$ and $dx=x du$ where $x=e^u$

When $x=0, u=-\infty$ and when $x=1, u=0$

(2) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{nu}u^ne^u du\end{equation*}$

(3) $\begin{equation*}\int_0^1 x^n(ln(x))^n dx=\int_{-\infty}^0 e^{(n+1)u}u^n du\end{equation*}$

Integrate by parts using the tabular method.

Sign	Differentiate	Integrate
+	$u^n$	$e^{(n+1)u}$
–	$nu^{n-1}$	$\frac{e^{(n+1)u}}{n+1}$
+	$n(n-1)u^{n-2}$	$\frac{e^{(n+1)u}}{(n+1)^2}$
–	$n(n-1)(n-2)u^{n-3}$	$\frac{e^{(n+1)u}}{(n+1)^3}$
+	$\frac{n!}{(n-4)!}u^{n-4}$	$\frac{e^{(n+1)u}}{(n+1)^4}$
	$\vdots$	$\vdots$
	$\frac{n!}{(n-n)!}u^{n-n}$	$\frac{e^{(n+1)u}}{(n+1)^{n}}$
$(-1)^n$	$0$	$\frac{e^{(n+1)u}}{(n+1)^{n+1}}$

When we substitute $u=-\infty$ or $u=0$ the differentiation column is zero except for $\frac{n!}{(n-n)!}u^{n-n}$ , which is $n!$ ,

Thus $\int_0^1 x^n(ln(x))^n dx=\frac{e^{(n+1)u}}{(n+1)^{n+1}}}]_{-\infty}^0$

$\begin{equation*}=n!\times\frac{e^0}{(n+1)^{n+1}}-0\end{equation}$

$\begin{equation*}=\frac{n!}{(n+1)^{n+1}}\end{equation}$

Now we just need to think about the sign.

$\begin{equation*}=(-1)^n\frac{n!}{(n+1)^{n+1}}\end{equation}$

The integral is now

$\int_0^1{x^x dx}=(\Sigma_{n=0}^{\infty}(\frac{1}{n!}( (-1)^n\frac{n!}{(n+1)^{n+1}})$

So $\int_0^1{x^x dx}=\Sigma_{n=0}^{\infty}( (-1)^n\frac{1}{(n+1)^{n+1}}$

Let’s work out some partial sums

$\int_0^1{x^x dx}=0.778343$

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Filed under Definite, Integration, Integration by Parts, Tabular Integration

Tagged as Integration, integration by parts, tabular integration

November 17, 2024 · 5:23 am

Integration by Parts using the Tabular Method

(1) $\begin{equation*}\int_0^1x^2e^xdx\end{equation*}$

I am going to do this integral in two ways; the traditional method and the tabular method.

Traditional Method

Remember $\int{u dv}=u\times v-\int{v du}$

Let $u=x^2$ and $dv=e^x$

Then $du=2x$ and $v=\int{e^x dx}=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-\int_0^1{e^x 2x dx}\end{equation}$

Now we need to do integration by parts on $\int_0^1{e^x 2x dx}$

Let $u=2x$ and $dv=e^x$

Then $du=2$ and $v=e^x$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-\int_0^1{e^x 2 dx})\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-(2e^x )]_0^1)\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e-(2e-(2e-2))\end{equation}$

$\begin{equation*}\int_0^1{x^2e^x dx}=e+2\end{equation}$

Tabular Integration

Similar to before, select a $u$ and a $dv$ , $u=x^2$ and $dv=e^x$

Sign	D(ifferentiate)	I(ntegrate)
+	$x^2$	$e^x$
–	$2x$	$e^x$
+	$2$	$e^x$
–	$0$	$e^x$

Stop when the differentiating column reaches zero.

Then we multiply diagonally

$(+x^2)(e^x)+(-2x)(e^x)+(+2e^x)$

$=x^2e^x-2xe^x+2e^x]_0^1$

$=e-2e+2e-(0-0-2)$

$=e+2$

It is only worth using this method if integration by parts is required more than once. Also, the $u$ has to eventually differentiate to $0$ .

Let’s try another one

(2) $\begin{equation*}\int{x^3cos(2x) dx}\end{equation*}$

Let $u=x^3$ and $dv=cos(2x)$

Sign	D	I
+	$x^3$	$cos(2x)$
–	$3x^2$	$\frac{1}{2}sin(2x)$
+	$6x$	$-\frac{1}{4}cos(2x)$
–	$6$	$-\frac{1}{8}sin(2x)$
+	$0$	$\frac{1}{16}cos(2x)$

$\int{x^3cos(2x) dx}=(x^3)(\frac{1}{2}sin(2x))+(-3x^2)(-\frac{1}{4}cos(2x))+(6x)(-\frac{1}{8}sin(2x))+(-6)(\frac{1}{16}cos(2x))+c$

$=\frac{x^3}{2}sin(2x)+\frac{3x^2}{4}cos(2x)-\frac{3x}{4}sin(2x)-\frac{3}{8}cos(2x)+c$

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Filed under Integration, Integration by Parts, Tabular Integration

Tagged as Calculus, integration by parts, tabular integration

November 2, 2024 · 11:45 am

Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that $f(x)$ is continuous everywhere and that $\int_{4}^{10} f(x) dx=-10$ , find:

(a) $\int_{4}^{10}2x +f(x) dx$

(b) $\int_{5}^{11} f(x-1) dx$

(c) $\int_{1}^{3} f(3x+1) dx$

(d) $\int_{-10}^{-4} -f(-x) dx$

(e) $\int{10}^{22} f(\frac{x-2}{2}) dx$

(f) $\int_{-3}^{-9} f(1-x) dx$

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx

Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).

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Filed under Definite, Integration, Uncategorized, Year 12 Mathematical Methods

June 12, 2024 · 7:02 am

The Logistic Equation

My year 12 Specialist students are working on logistic growth at the moment. An example might be helpful.

A new viral disease was found to spread according to the equation $\frac{dN}{dt}=kn(M-N)$ , where $M$ is the susceptible population, $N$ is the number of people infected at time $t$ months and $k=1.5\times 10^{-9}$ . In March 2010, it was thought only 100 people out of a population of 18 million were infected. Use the logistic model to find the number infected in:

(a) March 2011

(b) June 2012

(c) January 2017
Specialist 12 – Nelson Senior Maths