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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: squance on May 08, 2009, 08:57:12 pm

Title: Antidifferentiation Questions
Post by: squance on May 08, 2009, 08:57:12 pm: My sis would like some help with some past spesh exam questions:

$\int {f(x)dx=2}dx, then \int{(2f(x)+3)dx$ is equal to? (both upper and lower bounds of both intergrals are 4 and 1 respectively).

A. 2
B. 4
C. 7 (my sis reckons its this answer but not sure).
D. 10
E. 13

There was this question where she had to find d/dx(x cos x) and she did : the answer she got was -xsinx + cosx
Then the next part of the question was:
Hence show that an antiderivative of x sin x is (sinx -x cosx) . (with me glancing quickly at this question, its integration by parts yeah? and are year 12's expected to know how to integrate by parts?)

And then there was this question where she had to sketch f(x) = x^3-25/5x. She can't seem to do it on her graphics calculator and wants to know what the graph looks like.

Help muchly appreciated
Title: Re: Antidifferentiation Questions
Post by: Mao on May 08, 2009, 09:03:47 pm: Code: [Select]
\int_{1}^{4} f(x)\; dx
$\int_1^4 (2f(x)+3)\; dx = 2\int_1^4 f(x)\; dx + \int_1^4 3 dx = 2 \times 2 + 9 = 13$ unit²

The second question is integration by recognition:
$\begin{align*} \frac{d}{dx} (x\cdot \cos(x)) & = -x\cdot \sin(x) + \cos(x) \\ \implies x\cdot \cos(x) + C & = \int -x \cdot \sin(x) + \cos(x) \; dx\\ x\cdot \cos(x) + C & = -\int x\cdot \sin(x)\; dx + \int \cos(x)\; dx \\ \int x\cdot \sin(x)\; dx & = \int \cos(x)\; dx - x\cdot \cos(x) \\ \int x\cdot \sin(x)\; dx & = \sin(x) - x\cdot \cos(x) + C\\ \end{align*}$

Shown as required.
Title: Re: Antidifferentiation Questions
Post by: squance on May 08, 2009, 09:23:11 pm: Thanks Mao
Title: Re: Antidifferentiation Questions
Post by: TrueTears on May 08, 2009, 09:44:45 pm: hey TinyApple do you mean (x^3-25)/5x or x^3 - (25/5x)?

assuming (x^3-25)/5x

Long dividing yields : $\frac{1}{5}x^2 - \frac{5}{x}$

Now as $x \to \pm \infty$ , $\frac{5}{x} \to 0$ so we have an asymptote $y = \frac{1}{x}x^2$

also $x \neq 0$ , so there is a vertical asymptote at x = 0

now we have $y = \frac{1}{5}x^2 + \frac{-5}{x}$ to sketch using addition of ordinates.

(http://img413.imageshack.us/img413/9425/tinyapple.jpg)

Worst paint ever lol, but yeah you literally just 'add' the y coordinates of the 2 graphs to get the resultant graph, hence the name addition of ordinates.
Title: Re: Antidifferentiation Questions
Post by: squance on May 08, 2009, 10:07:24 pm: ! :)

Thanks Truetears.

My sis also wants to know how to find the turning point of the parabola (in the 2nd quadrant)..if possible?
Title: Re: Antidifferentiation Questions
Post by: TrueTears on May 08, 2009, 10:21:11 pm: Ah that parabola-looking shape

So you will need to find $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{2x}{5} + \frac{5}{x^2}$

Now we require $\frac{dy}{dx} = 0$

$\frac{2x}{5} = -\frac{5}{x^2}$

Cross multiply leads $x^3 = \frac{25}{-2}$

$x = \frac{-10^{\frac{2}{3}}}{2}$

Subbing this value of x in and you can work out the y value and hence get the coordinate $(x,y)$
Title: Re: Antidifferentiation Questions
Post by: squance on May 09, 2009, 09:31:01 am: THanks again Truetears.
My sis greatly appreciates it :)
Title: Re: Antidifferentiation Questions
Post by: Juddinator on March 07, 2010, 12:16:28 pm: I seem to be having problems differentiating cos(4x) with limits pie/2 and 0. The answer I get is 1 however the back of the book says 0. I antidifferentiate cos to sin and go from there but my problems continue...
Title: Re: Antidifferentiation Questions
Post by: kyzoo on March 07, 2010, 12:43:38 pm: Antiderivative of cos(4x) = 0.25sin(4x)

So if you have limits pi/2 and 0

Value of definite integral = 0.25sin (2pi) - 0.25 sin (0) = 0.25 sin (0) - 0.25 sin (0) = 0
Title: Re: Antidifferentiation Questions
Post by: m@tty on March 07, 2010, 12:45:10 pm: Consider $cos(4x)$ graph, its period is $\frac{2\pi}{4} = \frac{\pi}{2}$ . Hence it completes one entire period over the interval $[0,\frac{\pi}{2})$ , by symmetry, the area above the x-axis and that below the x-axis are equal in magnitude. So the 'unsigned' area will be zero. But by splitting the integral in half you can determine the actual area. ie. $2 \cdot\int_{0}^{\frac{\pi}{4}} \cos{4x} \ dx$

You probably know the above already :buck2:

They would've wanted you to determine the value of the integral $\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx$ , which is zero. Rather than find the area between the curve and the x-axis.

$\begin{align}\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx =& \frac{1}{4}[\sin{4x}]_{0}^{\frac{\pi}{2}} \\ =& \frac{1}{4}(0-0) \\ =&0 \ units^2\end{align}$
Title: Re: Antidifferentiation Questions
Post by: Juddinator on March 07, 2010, 01:08:05 pm: Quote from: kyzoo on March 07, 2010, 12:43:38 pm
Antiderivative of cos(4x) = 0.25sin(4x)

So if you have limits pi/2 and 0

Value of definite integral = 0.25sin (2pi) - 0.25 sin (0) = 0.25 sin (0) - 0.25 sin (0) = 0
Quote from: m@tty on March 07, 2010, 12:45:10 pm
Consider $cos(4x)$ graph, its period is $\frac{2\pi}{4} = \frac{\pi}{2}$ . Hence it completes one entire period over the interval $[0,\frac{\pi}{2})$ , by symmetry, the area above the x-axis and that below the x-axis are equal in magnitude. So the 'unsigned' area will be zero. But by splitting the integral in half you can determine the actual area. ie. $2 \cdot\int_{0}^{\frac{\pi}{4}} \cos{4x} \ dx$

You probably know the above already :buck2:

They would've wanted you to determine the value of the integral $\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx$ , which is zero. Rather than find the area between the curve and the x-axis.

$\begin{align}\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx =& \frac{1}{4}[\sin{4x}]_{0}^{\frac{\pi}{2}} \\ =& \frac{1}{4}(0-0) \\ =&0 \ units^2\end{align}$
Yeah that makes much more sense guys, thanks a heap :D. It only makes sense that by splitting the integral in half you can determine the actual area XD.

Thanks again kyzoo and m@tty
Title: Re: Antidifferentiation Questions
Post by: moekamo on March 07, 2010, 11:00:41 pm: Quote from: m@tty on March 07, 2010, 12:45:10 pm
Consider $cos(4x)$ graph, its period is $\frac{2\pi}{4} = \frac{\pi}{2}$ . Hence it completes one entire period over the interval $[0,\frac{\pi}{2})$ , by symmetry, the area above the x-axis and that below the x-axis are equal in magnitude. So the 'unsigned' area will be zero. But by splitting the integral in half you can determine the actual area. ie. $2 \cdot\int_{0}^{\frac{\pi}{4}} \cos{4x} \ dx$

You probably know the above already :buck2:

They would've wanted you to determine the value of the integral $\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx$ , which is zero. Rather than find the area between the curve and the x-axis.

$\begin{align}\int_{0}^{\frac{\pi}{2}} \cos{4x} \ dx =& \frac{1}{4}[\sin{4x}]_{0}^{\frac{\pi}{2}} \\ =& \frac{1}{4}(0-0) \\ =&0 \ units^2\end{align}$

if the question is a pure definite integral with no mention of area under a curve then you do not need to put units squared on the end, its just a number.
Title: Re: Antidifferentiation Questions
Post by: Juddinator on March 08, 2010, 08:03:36 am: When doing antiderivatives involving inverse circular functions and we have $\frac{1}{\sqrt{5-x^2}}$ , shouldn't the 'a' term just come out to make it sin-1 (x/5)? What about $\frac{3}{16+x^2}$ ?

Thanks! :D (Sorry I couldn't get the integral sign going guys!)
Title: Re: Antidifferentiation Questions
Post by: moekamo on March 08, 2010, 08:42:31 am: Quote from: Juddinator on March 08, 2010, 08:03:36 am
When doing antiderivatives involving inverse circular functions and we have $\frac{1}{\sqrt{5-x^2}}$ , shouldn't the 'a' term just come out to make it sin-1 (x/5)? What about $\frac{3}{16+x^2}$ ?

Thanks! :D (Sorry I couldn't get the integral sign going guys!)

$\int \frac{1}{\sqrt{5-x^2}} dx = \int \frac{1}{\sqrt{(\sqrt{5})^2-x^2}} dx=\sin^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$

$\int \frac{3}{16+x^2} dx = 3 \int \frac{1}{4^2+x^2} dx=\frac{3}{4}\tan ^{-1}\left ( \frac{x}{4}\right) +c$

the integral sign in latex is \int
Title: Re: Antidifferentiation Questions
Post by: Juddinator on March 08, 2010, 01:48:02 pm: Quote from: moekamo on March 08, 2010, 08:42:31 am
Quote from: Juddinator on March 08, 2010, 08:03:36 am
When doing antiderivatives involving inverse circular functions and we have $\frac{1}{\sqrt{5-x^2}}$ , shouldn't the 'a' term just come out to make it sin-1 (x/5)? What about $\frac{3}{16+x^2}$ ?

Thanks! :D (Sorry I couldn't get the integral sign going guys!)

$\int \frac{1}{\sqrt{5-x^2}} dx = \int \frac{1}{\sqrt{(\sqrt{5})^2-x^2}} dx=\sin^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$

$\int \frac{3}{16+x^2} dx = 3 \int \frac{1}{4^2+x^2} dx=\frac{3}{4}\tan ^{-1}\left ( \frac{x}{4}\right) +c$

the integral sign in latex is \int
Thanks Moekamo (Y)
Title: Re: Antidifferentiation Questions
Post by: superflya on March 08, 2010, 02:02:48 pm: remember its in the form $a^{2}$ and becomes $a$ after integrating.
Title: Re: Antidifferentiation Questions
Post by: Juddinator on March 08, 2010, 06:01:18 pm: Quote from: superflya on March 08, 2010, 02:02:48 pm
remember its in the form $a^{2}$ and becomes $a$ after integrating.
yes, true. thanks superflya