July 08, 2020, 02:33:38 pm

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#### Einstein_Reborn_97

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« Reply #4470 on: June 02, 2020, 10:35:26 pm »
+1
so in other words, its a limiting sum because its height could go on infinitely?
Not exactly. When -1<r<1, the sum of a geometric series does not continue increasing greatly after a few times, instead it approaches some constant value - the limiting sum. E.g. 8+4+2+1+..., this series is said to converge, you can see that as the series continues, each term gets smaller and smaller (approaching zero) and eventually any changes to the sum of the series will be infinitesimal.
Contrast this with the series 2+4+8+16+32+..., the sum will become very large as n increases and we say these series diverge (their sum is infinite).

Hope that clarifies everything
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#### RuiAce

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« Reply #4471 on: June 05, 2020, 11:25:37 am »
0
If a question says "give an answer correct to ... decimal places" how do you know if you should round? From what I've seen, some answers are rounded and some aren't.
If they're not rounding correctly then that's incorrect. You should always be rounding.

#### Einstein_Reborn_97

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« Reply #4472 on: June 05, 2020, 11:26:59 am »
0
If a question says "give an answer correct to ... decimal places" how do you know if you should round? From what I've seen, some answers are rounded and some aren't.
Always round off to whatever number of decimal places the question specifies...that's the correct way to present your answer. Don't worry if the answers from a textbook are different from yours by a single digit at the end because of different rounding. Just check your working out and re-calculate the answer. If your process is correct, you wouldn't lose marks for that.
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#### Coolmate

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« Reply #4473 on: June 10, 2020, 11:09:23 pm »
0
Hey Everyone!

Could I please have some help with this Definite Integral question: $\int_{-4}^{1} x^2 dx$

I get the answer of: $20 \frac{5} {3}$

But the answer is meant to be: $\frac{2} {3}$

I have attached my full working, could someone please identify where I am going wrong?

Coolmate
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#### wsdm

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« Reply #4474 on: June 11, 2020, 12:12:31 am »
+3
I have attached my full working, could someone please identify where I am going wrong?
There doesn't seem to be any error in your working out. I also attained $\frac{65}{3}$ as my final answer. Are you completely sure there wasn't any typo in the question? If not, the textbook (if you're getting this off a textbook) might have the wrong answer, but don't worry, that happens a lot.
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#### fun_jirachi

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« Reply #4475 on: June 11, 2020, 12:44:02 pm »
+3
Hey Everyone!

Could I please have some help with this Definite Integral question: $\int_{-4}^{1} x^2 dx$

I get the answer of: $20 \frac{5} {3}$

But the answer is meant to be: $\frac{2} {3}$

I have attached my full working, could someone please identify where I am going wrong?

Coolmate

Your working is totally fine - the only way I think $\frac{2}{3}$ could be a reasonable answer is if we had the integral $\int_{-1}^1 x^2 \ dx$. Perhaps you copied the bounds wrong (highly unlikely, but possible) or more likely the answers are just plain wrong.

Note that it's often redundant to rewrite your answer as a mixed fraction/decimal as the improper fraction is sufficient - leave it as an improper fraction (it'll save you time and computation!). Depending on how pedantic your teachers are, you may want to add $u^2$ as units as well.
« Last Edit: June 11, 2020, 02:19:56 pm by fun_jirachi »
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#### Coolmate

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« Reply #4476 on: June 11, 2020, 06:13:04 pm »
+3
There doesn't seem to be any error in your working out. I also attained $\frac{65}{3}$ as my final answer. Are you completely sure there wasn't any typo in the question? If not, the textbook (if you're getting this off a textbook) might have the wrong answer, but don't worry, that happens a lot.

Your working is totally fine - the only way I think $\frac{2}{3}$ could be a reasonable answer is if we had the integral $\int_{-1}^1 x^2 \ dx$. Perhaps you copied the bounds wrong (highly unlikely, but possible) or more likely the answers are just plain wrong.

Note that it's often redundant to rewrite your answer as a mixed fraction/decimal as the improper fraction is sufficient - leave it as an improper fraction (it'll save you time and computation!). Depending on how pedantic your teachers are, you may want to add $u^2$ as units as well.

Thankyou wsdm and fun_jirachi, for the assistance!

So, funny story, I asked my teacher about it, then she looked at the question and what I wrote, and she showed me how I copied the bounds from another question, but wrote the "x" value from the right question..... . So you were correct fun_jirachi, I did copy the bounds wrong.

Thankyou again for the help!
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#### twelftholmes

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« Reply #4477 on: June 22, 2020, 04:16:39 pm »
0
Hey! This is a HSC question from the 2010 paper (5b)

(the question is attached as a jpg if you don't mind downloading it, I'm not sure how you insert a picture properly into a post oop)

I understand proving parts (i) and (ii) but I don't understand how to connect it to part (iii). I looked at the worked solutions but I can't make sense of it, could someone please explain it to me like I'm 5? haha thanks in advance!

#### fun_jirachi

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« Reply #4478 on: June 22, 2020, 04:45:09 pm »
+2
Hey there!

You can use an image hosting service like imgur to include photos - if you want more information or are unsure how to do it, there's plenty of info around the forum/ you can ask again!

It's important to note that the question is a hence question. This implies you have to use the previous parts at some point or other. While this is usually as early as possible in the question, it very occasionally isn't - but luckily, here it definitely is. Thus, we sub in what we just proved in part (ii) to conclude that
$\int_0^\frac{\pi}{4} \frac{dx}{1-\sin x} = \int_0^\frac{\pi}{4} (\sec ^2x + \sec x \tan x) \ dx$

From here, it's important to recognise that $\int \sec ^2x \ dx$ is a standard integral that appears on your reference sheet - and that $\int \sec x \tan x \ dx$ is given to you as part of the question. Hence, we find that

$\int_0^\frac{\pi}{4} (\sec ^2x + \sec x \tan x) \ dx = \left[\tan x + \sec x\right]_0^\frac{\pi}{4} = \sqrt{2}$

Hope this helps
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#### twelftholmes

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« Reply #4479 on: June 22, 2020, 06:17:10 pm »
0
Thank you so much! The way you explained it is extremely easy to understand.
Also thanks for telling me I could use Imagur, I went and found out how to do it under New Users Lounge and its much better

If you don't mind there's this other question that I tried to work out myself but I have no idea where to start, could you please help me understand it? It's not from any HSC paper so I can't find worked solutions online, but I think it's from a bank where teachers have access to HSC style questions.

https://imgur.com/a/nFPINIq

Thank you so much in advance!
« Last Edit: June 22, 2020, 06:21:14 pm by twelftholmes »

#### fun_jirachi

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« Reply #4480 on: June 22, 2020, 06:40:25 pm »
+1
Usually, the first thing that should be tried with composite area is to split it up. We can see that we have the areas defined by $\int_0^1 \sin (\pi x) \ dx$ and $\int_1^a \sin (\pi x) \ dx - a \times \sin (a\pi)$. Note that the former comes from the area enclosed by the sine curve and the x-axis, while the second area is defined by the rest of the shaded area. It's also important to note that we're looking at signed area - so the latter area's signs are very important (we're looking at what is essentially a more abstract primitive function, don't mix the signs up!). Even if you do mix the signs up, you'll probably notice because your answer will be different to the result you're asked to prove, but the justification is here for reference.

From here, you can integrate to find $A(a)$ in simpler terms. Answer in the spoiler, but try and avoid using it where possible - I think it's important to get as much practice as possible with questions of this sort
Spoiler
\begin{align*} A(a)
&= \int_0^1 \sin (\pi x) \ dx + \int_1^a \sin (\pi x) \ dx - a \times \sin (a\pi)
\\&= \left[-\frac{1}{\pi}\cos (\pi x)\right]_0^a - a\times \sin (a\pi)
\\&=\frac{1}{\pi} (1-\cos(a\pi)) - a\sin (a\pi)
\end{align*}
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
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ATAR: 99.05

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