Author Topic: 4U Maths Question Thread (Read 665492 times) Tweet Share

RuiAce · « **Reply #180 on:** May 03, 2016, 09:30:51 am »

Quote from: amandali on May 03, 2016, 01:39:15 am

how to solve
integral of sinx /[ 2+cosx - cos^2(x) ] and integral of 1/(e^2x + 4e^x + 9)

Doing these questions by hand today cause I'm not at home and typing LaTeX.

For Q1, whilst normally I advocate reverse chain rule, when it collapses into partial fractions it's easier to start on a substitution.

RuiAce · « **Reply #181 on:** May 03, 2016, 09:47:24 am »

The substitution isn't directly obvious for the second one, but the neat thing about exponentials is that you can "introduce" the substitution by multiplying top and bottom by e^x.

The final answer is subject to inaccuracy

birdwing341 · « **Reply #182 on:** May 03, 2016, 11:13:48 am »

Quote from: RuiAce on May 03, 2016, 09:47:24 am

The substitution isn't directly obvious for the second one, but the neat thing about exponentials is that you can "introduce" the substitution by multiplying top and bottom by e^x.

The final answer is subject to inaccuracy(Image removed from quote.)(Image removed from quote.)

Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2

jamonwindeyer · « **Reply #183 on:** May 03, 2016, 11:19:57 am »

Quote from: birdwing341 on May 03, 2016, 11:13:48 am

Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2

That's correct. That last inverse tangent term should be:

$-\frac{2}{9\sqrt{5}}\tan^{-1}{\frac{e^x+2}{\sqrt{5}}}$

Nasty integral

jakesilove · « **Reply #184 on:** May 03, 2016, 11:48:49 am »

Quote from: jamonwindeyer on May 03, 2016, 11:19:57 am

That's correct. That last inverse tangent term should be:

$-\frac{2}{9\sqrt{5}}\tan^{-1}{\frac{e^x+2}{\sqrt{5}}}$

Nasty integral

Really, really nasty. Had to use a program to get the correct answer, and I really can't imagine that you would be asked this in an exam situation (maybe as a final questions?). Ah well, I guess this is why every other State shakes in fear when they hear that we have an Extension 2 Maths course.

Jake

RuiAce · « **Reply #185 on:** May 03, 2016, 01:01:11 pm »

Quote from: birdwing341 on May 03, 2016, 11:13:48 am

Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2

Like I said, subject to inaccuracy. I paused the question and got back to it so I dropped 1/9 everywhere. Same goes with the completing the square dropoff

katherine123 · « **Reply #186 on:** May 03, 2016, 11:01:17 pm »

integral of 1/(x^2-9)^3/2

I let x=3sec(B)

and im left with
integral of 1/9*cos(B)*cosec^2(B)

RuiAce · « **Reply #187 on:** May 04, 2016, 12:15:48 am »

Quote from: katherine123 on May 03, 2016, 11:01:17 pm

integral of 1/(x^2-9)^3/2

I let x=3sec(B)

and im left with
integral of 1/9*cos(B)*cosec^2(B)

$\text{Rather than leave it as }\frac{1}{9}\int{\cos{\theta}\csc^2{\theta}\,d\theta}\\\text{Try writing it as }\frac{1}{9}\int{\frac{\cos{\theta}}{\sin^2{\theta}}d\theta}$
$\text{The reverse chain rule or substitution }u=\sin{\theta}\text{ should be much more obvious now.}$

$\text{Alternatively, more favourably observe that the integral can be rewritten as }\int{\csc{\theta}\cot{\theta}\,d\theta}\\\text{which automatically integrates into }-\csc{\theta}+C$

katherine123 · « **Reply #188 on:** May 05, 2016, 12:06:17 am »

how to solve
integral of 1/[x*(x^2+1)^1/2] with limits between 1 and 2?

RuiAce · « **Reply #189 on:** May 05, 2016, 12:24:36 am »

Quote from: katherine123 on May 05, 2016, 12:06:17 am

how to solve
integral of 1/[x*(x^2+1)^1/2] with limits between 1 and 2?

$\text{Once again we can apply a substitution to help out. But choose the better one.}\\ u^2=x^2+1\Rightarrow 2u\,du=2x\,dx \Leftrightarrow u\,du=x\,dx \quad (u>0)\\ x=1\Rightarrow u=\sqrt{2}\\ x=2\Rightarrow u=\sqrt{5}$
$\begin{align*}\int_1^2{\frac{dx}{x\sqrt{x^2+1}}}&=\int_1^2{\frac{x\,dx}{x^2\sqrt{x^2+1}}}\\&=\int_\sqrt{2}^\sqrt{5}{\frac{u\,du}{\left(u^2-1\right) u}}\\&=\int_\sqrt{2}^\sqrt{5}{\frac{du}{(u+1)(u-1)}}\end{align*}$
$\text{The rest is easy}$

Note: In the absence of the boundaries, to avoid the partial fractions it may have been more beneficial to use the trigonometric substitution x=tan(θ). However, dealing with boundaries that have inverse trigonometric functions in them, whilst not necessarily bad, can get messy because you don't know what the final result is.

aoifera · « **Reply #190 on:** May 05, 2016, 06:33:15 pm »

Hello,
I'm supposed to do this question with both washers and cylinders but I can't get either to work
Find the volume of a torus formed with the circle (x-2)² + y² = 1 is rotated about the y axis
Thank you

RuiAce · « **Reply #191 on:** May 05, 2016, 10:02:20 pm »

Quote from: aoifera on May 05, 2016, 06:33:15 pm

Hello,
I'm supposed to do this question with both washers and cylinders but I can't get either to work
Find the volume of a torus formed with the circle (x-2)² + y² = 1 is rotated about the y axis
Thank you

$\text{The shells method is quite easy.}\\\text{Take a typical shell and observe immediately that it's radius is dependent on }x\text{, but then that the height is based on double the value for }y.$
$\delta V = 2\pi rh \delta x= 2\pi * x * 2\sqrt{1-{(x-2)}^{2}}*\delta x$
$\text{Then proceed as normal}\\ \therefore V = \lim_{\delta x \rightarrow 0}{\sum_{x=1}^{3}{2\pi x \cdot 2\sqrt{1-{(x-2)}^{2}} \delta x }}\\ \text{I won't try to evaluate the integral myself but I have a feeling that after the substitution} \\ u=x-2\\ \text{you can use properties of odd/even functions. This comes out of memory and may be a bit unreliable.}$
$\textbf{If my memory is correct though, the odd function vanishes whereas the even function is }\frac{1}{4}\text{ the area of a circle}$

$\text{To take a typical washer however, we need to carefully consider what the outer and inner radii are.}\\ \text{Firstly, the equation rearranges into }x=2 \pm \sqrt{1-y^2} \\ \text{If you split up the circle properly, the right part of the circle contains the }+\\\text{and the left part of the circle contains the }-$

$\text{Take note that then the inner radius is}\\ r=2-\left(2-\sqrt{1-y^2}\right)=\sqrt{1-y^2}\\\text{And that the outer radius is}\\R= 2+\left(2-\sqrt{1-y^2}\right)=4-\sqrt{1-y^2}$
$\text{Rather than this formula }\delta V = \left(R^2-r^2\right)\delta y\\ \text{We will use difference of two squares to aid us.}\\\delta V = (R+r)(R-r)\delta y = (4)(4-2\sqrt{1-y^2})\delta y$
$\text{Then do the same systematic approach.}$

amandali · « **Reply #192 on:** May 07, 2016, 10:09:58 pm »

how do i integrate x^4*e^(-x)

jamonwindeyer · « **Reply #193 on:** May 08, 2016, 12:03:10 am »

Quote from: amandali on May 07, 2016, 10:09:58 pm

how do i integrate x^4*e^(-x)

Hey Amandali! You are looking at integration by parts for this question, but it is actually quite difficult, as it involves Recurrence Relations!

$\int uv' = uv - \int u'v$

So, the working would look a little like this.

$u = x^4 \\ u'=4x^3 \\ v' = e^{-x} \\ v=-e^{-x}$

$\int x^4e^{-x}dx = -x^4e^{-x}+4\int x^3e^{-x}dx$

The last part of that line should look familiar, it is the same integral we just performed! The only difference is the power has been reduced by one. This might be enough for you to spot the recurrence relation, but just in case, let's do that last integral by integration by parts.

$\int x^3e^{-x}dx = -x^3e^{-x}+3\int x^2e^{-x}dx$

You should start to see a pattern emerging here. The recurrence relation we are seeing is:

$I_n=-x^ne^{-x}+nI_{n-1}$

We can use this relation all the way to its conclusion (it breaks once we hit n=1), and find the answer. Be careful to carry the multiples all the way through each recurrence!

$I = -x^4e^{-x}+4(-x^3e^{-x}+3(-x^2e^{-x}+2(-xe^{-x}-e^{-x})))$

$I =-e^{-x}(x^4+4x^3+12x^2+24x+24)$

I hope this helps! Recurrence relations can be a little strange if you haven't seen them before, so let me know if anything is a little unclear!!

RuiAce · « **Reply #194 on:** May 08, 2016, 05:54:34 am »

Quote from: amandali on May 07, 2016, 10:09:58 pm

how do i integrate x^4*e^(-x)

$\text{Whilst Jamon basically showed the method of recurrence relations (aka. reduction formulae), here's the more direct approach.}$

$\begin{align*}\text{Let }I_n&=\int{x^n e^{-x} \, dx}\\\text{Then }I_n&=x^n \left(-e^{-x}\right)-\int{nx^{n-1} \left(-e^{-x}\right) \, dx}\text{ via IBP}\\&=-x^n e^{-x}+n\int{x^{n-1} e^x \, dx}\\&= -x^n e^{-x} + n I_{n-1} \end{align*}$

You may choose now to work backwards, or forwards. Here is how to work forwards.

$\begin{align*}\therefore \int{x^4 e^{-x} \, dx}&=I_4\\&=-x^4 e^{-x}+4 I_3\\&= -x^4 e^{-x}+4\left(-x^3 e^{-x}+3I_2 \right)\\&=-x^4 e^{-x} - 4x^3 e^{-x} + 12\left(-x^2 e^{-x} + 2I_1\right)\\ &= -x^4 e^{-x} - 4x^3 e^{-x} -12x^2 e^{-x} + 24\left(-xe^{-x} + I_0\right)\\ &= -x^4 e^{-x} - 4x^3 e^{-x} -12x^2 e^{-x}-24x e^{-x} + 24 \int{x^0e^{-x}\,dx}\\&= -x^4 e^{-x} - 4x^3 e^{-x} -12x^2 e^{-x}-24x e^{-x} -24 e^{-x} + C\end{align*}$

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