Category Archives: Integration

January 1, 2025 · 9:40 am

Volume of Rotation About a Slanting Line

Given the area in the first quadrant bounded by $x^2=12y$ , the line $y=3$ and the $y-$ axis. What is the volume generated when this area is rotated about the line $2x-y+4=0$ ?

Rotate the green region about the line $y=2x+4$

Rotate the green region about the line $y=2x+4$

We can split the solid into shells.

$\begin{equation*}V=2\pi r dx dy\end{equation}$

Where $r$ is the distance from each $(x,y)$ point in the region to the line $2x-y+4=0$ , $dx$ is the width, and $dy$ is the height.

The distance between a point and a line is

$\begin{equation*} d=\frac{Ax+By+C=0}{\sqrt{A^2+B^2}}\end{equation}$

Hence, $r=\frac{2x-y+4}{\sqrt{5}}$

$\begin{equation*}V=2\pi\int \int \frac{2x-y+4}{\sqrt{5}} dx dy\end{equation}$

Now we just need to work out the bounds.

$0\le y \le 3$ and $0\le x\le \sqrt{12y}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}} \int_0^3 \int_0^{\sqrt{12y}} 2x-y+4 dx dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 x^2-yx+4x]_0^{\sqrt{12y}} dy \end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 12y-\sqrt{12}y^{\frac{3}{2}+4\sqrt{12y} dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(6y^2-\frac{2\sqrt{12}}{5}y^{\frac{5}{2}}+\frac{8\sqrt{12}}{3}y^{\frac{3}{2}}]_0^3\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{2\sqrt{12}}{5}(9\sqrt{3})+\frac{8\sqrt{12}}{3}(3\sqrt{3}))\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{108}{5}+48)\end{equation}$

$\begin{equation*}V=\frac{804\pi}{\sqrt{5}}\end{equation}$

If we rationalise the denominator

$\begin{equation*}V=\frac{804\sqrt{5}\pi}{25}\end{equation}$

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Filed under Integration, Volume of Revolution

Tagged as Volume of revolution, volume of revolution about a slanted line

December 30, 2024 · 8:16 am

Volume of revolution about a line that is not an axis

Find the volume of the solid of revolution obtained by rotating the region bounded by $f(x)=x^3+1, g(x)=x^2, 0\le x\le 1$ about the line $y=3$ .

Rotate the green region about the line $y=3$

Washer Method

$\begin{equation*}V=\pi \int [f(x)]^2 dx \end{equation}$

The volume of the solid is the volume of $y=x^2$ rotated about $y=3$ subtract the volume of $y=x^3+1$ rotated about $y=3$ .

$\begin{equation*}V=\pi \int_0^1((3-x^2)^2-(3-(x^3+1))^2 dx\end{equation}$

$3-x^2$ is the distance (i.e radius) of the curve and the line.

$\begin{equation*}V=\pi \int_0^1(9-6x^2+x^4-(4-4x^3+x^6)) dx\end{equation}$

$\begin{equation*}V=\pi \int_0^1(5-6x^2+4x^3+x^4-x^6) dx\end{equation}$

$\begin{equation*}V=\pi (5x-2x^3+x^4+\frac{x^5}{5}-\frac{x^7}{7}]_0^1\end{equation}$

$\begin{equation*}V=\pi (5-2+1+\frac{1}{5}-\frac{1}{7})\end{equation}$

$\begin{equation*}V=\frac{142 \pi}{35}\end{equation}$

Shell Method

The shell method is much harder because we need to split the integral into two parts.

We need to rotate the green region about $y=3$ and the red region

$\begin{equation*}V=2\pi\int (xf(x))dx\end{equation}$

$\begin{equation*}V=2\pi[\int_0^1(3-y)\sqrt{y} dy+\int_1^2 (3-y)(y-1)^{\frac{1}{3}} dy]\end{equation}$

$3-y$ is the distance between each $y-$ value and the line of rotation. For example, if we were rotating about the $x-$ axis, the distance is $y$ .

$\sqrt{y}$ is the height of the cylinder between $0$ and $1$ . $1-(y-1)^{\frac{1}{3}}$ is the height of the cylinder between $1$ and $2$ . Refer back to Shell method for more information.

I used a calculator to find this integral

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Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

Tagged as Disc method, Volume of revolution, volume of revolution about a line, washer method

December 28, 2024 · 7:12 am

Volume of Revolution Method Two (Shell Method)

I am going to use the same example as I did for Method One (Disc or Washer Method).

If we rotate the shaded region about the $x-$ axis, we get an open hollow cylinder (like a pipe).

The width of the integral is $\delta y$ and the midpoint is $y$ .

The height of the cylinder is $x$ , but we need it in terms of $y$ , hence $x=f(y)$

The volume of the hollow cylinder is the volume of the outer cylinder subtract the volume of the inner cylinder.

$\begin{equation*}V=\pi (y+\frac{\delta y}{2})^2f(y)-\pi (y-\frac{\delta y}{2})^2 f(y)\end{equation}$

$\begin{equation*}V=\pi f(y)((y+\frac{\delta y}{2})^2-(y-\frac{\delta y}{2})^2)\end{equation}$

Which we can expand using a difference of squares.

$\begin{equation*}V=\pi f(y)(y+\frac{\delta y}{2}+y-\frac{\delta y}{2})(y+\frac{\delta y}{2}-y+\frac{\delta y}{2})\end{equation}$

$\begin{equation*}V=\pi f(y)(2y \delta y)\end{equation}$

$\begin{equation*}V=2\pi yf(y)\delta y\end{equation}$

The volume of the entire sold will be

$\begin{equation*}V=\Sigma_{y=a}^b 2 \pi yf(y)\delta y\end{equation}$

As $\delta y \rightarrow 0$

$\begin{equation*}V=\lim\limits_{\delta y \to 0}\Sigma_{y=a}^b 2 \pi yf(y)\delta y=\int_a^b 2\pi yf(y) dy\end{equation}$

Even though we are rotating the line about the $x-$ axis, we are integrating with respect to the $y-$ axis.

Example

Find the volume of the solid generated by revolving the region between $y=x^2$ and $y=2x$ about the $y-$ axis.

If we are rotating about the $y-$ axis, we will integrate with respect to $x$ .

$\begin{equation*}V=2\pi \int x f(x) dx\end{equation}$

The height of our hollow cylinder is $2x-x^2$

Hence

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

$\begin{equation*}V=2\pi \int_0^2 (2x^2-x^3) dx\end{equation}$

$\begin{equation*}V=2\pi (\frac{2x^3}{3}-\frac{x^4}{4}]_0^2\end{equation}$

$\begin{equation*}V=2\pi (\frac{2}{3}\times 8-\frac{1}{4}\times 16 )\end{equation}$

$\begin{equation*}V=32 \pi(\frac{1}{3}-\frac{1}{4})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

Let’s check with method one.

$x^2=y$ and $x=\frac{y}{2}$

$\begin{equation*}V=\pi \int_0^4 y-\frac{y^2}{4} dy\end{equation}$

$\begin{equation*}V=\pi (\frac{y^2}{2}-\frac{y^3}{12})]_0^4\end{equation}$

$\begin{equation*}V=\pi(8-\frac{16}{3})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

I try to pick the method that makes the integration easier.

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Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

Tagged as shell method, Volume of revolution

December 23, 2024 · 3:45 am

Volume of Revolution – Method One (Disc or Washer Method)

If we rotate this line segment around the $x-$ axis, we generate a three dimensional solid.

We are going to find the volume of this solid.

Consider a small section of the line segment and rotate this about the $x-$ axis.

As the width of the section $(\delta x)$ gets smaller (i.e. $\rightarrow 0$ ), the solid is a cylinder.

The radius of the cylinder is $f(x)$ and the height of the cylinder is $\delta x$ .

The volume of a cylinder is $V=\pi r^2 h$

Hence the volume of our section is

$\begin{equation*}V=\pi[f(x)]^2\delta x\end{equation}$

If we divide our line segment into a large number of cylinders (of equal height) then,

$\begin{equation*}V=\Sigma_a^b(\pi [f(x)]^2\delta x\end{equation}$

where $a$ is the lower $x$ value and $b$ the upper.

Now we want $\delta x\rightarrow 0$ so $V=\lim\limits_{\delta x \to 0} \Sigma_a^b(\pi [f(x)]^2\delta x$

Which is

$\begin{equation*}V=\int_a^b \pi [f(x)]^2 dx\end{equation}$

Example

The curve $y=\sqrt{x-1}$ , where $2\le x\le5$ is rotated about the $x-$ axis to form a solid of revolution. Find the volume of this solid.

$\begin{equation*}V=\pi \int_2^5( y^2 dx)\end{equation}$

$\begin{equation*}V=\pi \int_2^5 x-1 \space dx \end{equation}$

$\begin{equation*}V=\pi (\frac{x^2}{2}-x]_2^5)\end{equation}$

$\begin{equation*}V=\pi(\frac{25}{2}-5-(\frac{4}{2}-2))\end{equation}$

$\begin{equation*}V=\frac{15 \pi}{2}\end{equation}$

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Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

Tagged as Disc method, Integration, volume, Volume of revolution

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