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chocolatebanana
Chocolatebanana

secondary 4 chevron_right A Maths chevron_right Singapore

please help with q17 the integration part. i've alr solved partial frac, the form is 12/5(2x+3) -6/5(x-1) +1/(x-1)^2
please also help with q18 the whole qn. thankyou!!

Date Posted: 4 years ago
Views: 209
Eric Nicholas K
Eric Nicholas K
4 years ago
Will answer at 3 am, starting from scratch, if no one has already done so.
J
J
4 years ago
Clue :

The one with denominator (x-1)² can be integrated by working backwards. It might help to change it into the form (x-1)^(-2) first.

For the other two terms, they should be integrated back to ln∣ something ∣

Try to solve it first , if you're really stuck then refer to Eric's solutions later
Eric Nicholas K
Eric Nicholas K
4 years ago
I can’t remember if the modulus sign is learnt in O Levels. Ten years ago when I learnt it, I only saw the modulus in the A Levels.
J
J
4 years ago
Yes, same here. It wasn't covered in O's , only introduced in A's back then. Modulus is the more correct one. Chocolatebanana, if your teachers/school teaches to put in brackets instead, then use brackets as per normal
chocolatebanana
Chocolatebanana
4 years ago
thankyou eric and j!
i've tried the qn again, however i still got the wrong answer. and also, what do you mean by modulus or brackets? does modulus have anything to do with integration? because i've only heard of modulus graph :( or maybe like you said it's not tested or o's
J
J
4 years ago
Can show your answer here? And can I know what's the correct answer provided?

Oh don't worry, just use brackets for the ln. You will learn about using modulus for ln in integration for A levels (if you choose JC).

Your partial fractions is correct for 17a)
chocolatebanana
Chocolatebanana
4 years ago
ohh okay sure thankyou!
for 17, my final step was 12/10 (ln 45-ln35) - 6/5 (ln10-ln5) + 1 which is 0.46980.
but the correct answer is 0.0302
J
J
4 years ago
actually 12/10 can simplified to 6/5.

You should be getting :

6/5 (ln9 - ln7) + ½ - 6/5 ln2

= -0.0302(3s.f)


From your final step, it seems that you have a x5 in your ln terms, which is incorrect. I think you have put the ln on '5(x -1)' and '5(2x+3)' .

The 5 is part of the fraction, which is a constant. It's not part of the variable so we cannot include it into the term to be ln-ed
chocolatebanana
Chocolatebanana
4 years ago
ohh, i get it now but may i ask how did u integrate 1/(x-1)^2?
Eric Nicholas K
Eric Nicholas K
4 years ago
This we just raise the power by 1 as usual
Eric Nicholas K
Eric Nicholas K
4 years ago
The only time we do not raise the power by 1 is when it’s things like 1/x.
J
J
4 years ago
Change to (x - 1)^(-2) first.


So when you integrate,

① Increase the power by 1.
② divide the expression by the differentiation of (x - 1). It is actually just 1 so dividing by 1 is just getting back the same expression.

Edit :
③ divide it by the value of the power you changed to.
In this case , -2 was increased to -1. So divide it by -1. Effectively it just changes the expression to negative. You can usually combine steps ② and ③ in one step.

You end up with -(x-1)^(-1), which = -1/(x - 1)


Once you are familiar with these, you would be able to change the powers straight away and divide by the corresponding constants without having to convert to ^(- something)


Eg. When you see something like

5/(2x + 7)⁴, you can directly integrate to

get 5/((-3)(2)(2x+7)³)
Eric Nicholas K
Eric Nicholas K
4 years ago
You can integrate by this method as Long as the term inside the bracket is linear. In your case, x - 1 is linear. Things like (x^2 + 1)^-2, however, cannot be integrated the usual way.
chocolatebanana
Chocolatebanana
4 years ago
thankyou j and eric for your help!! i understand this question now! :)
chocolatebanana
Chocolatebanana
4 years ago
if it's possible, could you help me with q18 too? please take your time if you can help me! i can't find the partial fraction. my answer was -2/x + 4/x+4 + 4/x-4. but the ans is supposed to be -1/2x + 1/4(x-4) + 1/4(x-4)
Eric Nicholas K
Eric Nicholas K
4 years ago
Ok, will look at them later. Currently still on the way back as I just ended a class earlier.
chocolatebanana
Chocolatebanana
4 years ago
ok thankyou so much! do rest too:)
J
J
4 years ago
8/(x(x² - 16))

= 8/(x(x + 4)(x - 4))

You probably equated it to the form

A/x + B/(x+4) + C/(x-4)

So 8 = A(x+4)(x-4) + Bx(x-4) + Cx(x+4)


Let x = 0,

8 = A(0+4)(0-4)
8 = -16A
A = -8/16 = -½

I think for your terms, you might have accidentally divided -16 by 8 instead of the other way round. Once you rectify this, you should get the correct answer.
chocolatebanana
Chocolatebanana
4 years ago
ohh i did exactly that, it was the last step that i was careless in. i get it now thankyou so much!!
J
J
4 years ago
Welcome. The integration should be very easy now

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Eric Nicholas K
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1st
Q17

The final answer is negative. Do not remove the negative sign in this case, as the integrals are meant to be that way.

When we integrate in this way (the usual way of course), technically we take the total shaded area above the x-axis minus the total shaded area below the x-axis. Of course, if the majority of the shaded area lies below the x-axis, the value of the integral will be negative.

Also, for your syllabus, integrating 1/x gives you ln x. In A Levels, for completeness, the relevant integral includes a modulus sign, which in this case is ln |x|, to account for the fact that the graph of 1/x also exists for x < 0 (and hence the integral should have a value), whereas plain ln x is not defined (at least not a real number), for non-positive values of x.
chocolatebanana
Chocolatebanana
4 years ago
thankyou so much!
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Q18