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secondary 4 | A Maths
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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!!
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
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
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)
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
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
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)
= 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.
See 2 Answers
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.