Category Archives: Definite

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.

SignDifferentiateIntegrate
+u^ne^{(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)^n0\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

n((-1)^n\frac{1}{(n+1)^{n+1}})
5=0.78343
10=0.78343
20=0.78343
100=0.78343

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

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

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.

(a) \int_{4}^{10} 2x +f(x) dx
\int_{4}^{10} 2x  dx +\int_{2}^{10} f(x) dx
(x^2]_4^{10} + (-10)
10^2-4^2-10
=74

(b) \int_{5}^{11} f(x-1) dx
=-10




(c) \int_{1}^{3} f(3x+1) dx
Let u=3x+1
\frac{du}{dx}=3
dx=\frac{du}{3}

When x=1, u=4 and when x=3, u=10
\int_{4}^10 f(u) \frac{du}{3}
=\frac{1}{3}\times (-10)
=\frac{-10}{3}

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

Let u=-x
\frac{du}{dx}=-1
dx=-du

When x=-4, u=4 and when x=-10, u=10
-\int_{4}^{10} f(u) -dx
=-10

(e) \int_{10}^{22} f(\frac{x-2}{2} dx
Let u=f(\frac{x-2}{2})
\frac{du}{dx}=\frac{1}{2}
\du=2dx

When x=10, u=4 and when x=22, u=10
2\int_{4}^{10} f(u) du
=-20

(f) \int_{-3}^{-9} 2f(1-x) dx
Let u=1-x
\frac{du}{dx}=-1

When x=-3, u=4 and when x=-9, u=10
=-2\int_{4}^{10} f(u) du
=20


Split the integral
Integrate the first part.


This is a horizontal translation (one unit to the right) so the shape of the curve doesn’t change.
The integration bounds have also shifted one unit to the right.




This is a horizontal dilation and translation. The easiest method is to use a change of variable





































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