jinny1's Methods Questions Thread :D

[moekamo]

From Derrick Ha's book

Calculating an 'area between curves' is interesting in that it does not matter whether the area is
above, below or overlapping the axes. It is worthwhile justifying to yourself why this is the case.

Could someone explain why this is the case?

thanks

another way to think about it, you know that for an area between curves when they are both above the x axis is just the integral of 'top - bottom' between the terminals.

Now wherever they are, whether it is below and above, or only below, we can translate both curves up by the same amount by adding a constant, h, to each curve. Now when computing the integral we do 'top - bottom' and the +h from each curve cancels out. Therefore we can just do top - bottom for area between two curves for any of the situations mentioned.

[jinny1]

Hi guys just wondering about the amount working out i have to show in methods/spesh examination 2 (tech enabled)..

So lets say on one of the spesh/methods exam 2, therre was a question asking me to factorise 2x^2+4x+3.. Can i just input that into the calculator and just copy/paste thw answer straight from the calc?? Or do i need to say first " using the factorising function in the calc it was found Blah blah...) or would i need to show working out by using the "cross method" of factorisation. The latter  defeats the whole point oftech enabled exams though...as it would just be like sitting exam 1

Thanks

[pi]

I would just write the factorised form straight away. If you want, you could make an intermediate step, but I think it's not needed.

[jinny1]

Even if it was a 2~4 mark question???

and by intermediate step, do you mean word explanations or actual working out ??

thanks

[pi]

Even if it was a 2~4 mark question???

and by intermediate step, do you mean word explanations or actual working out ??

Working out (like show that you have steps in factorisation, eg. complete square, etc.), although as it is CAS, I doubt it'll be more than 1-2 marks(s) for something like that

[jinny1]

Please need help with Exam 1 vce 2010



i dont get b.

[b^3]

so for a you would have got 2x*ln(x)+x
so if you differentiate that you will get x²ln(x)
so you need to know the anitderivative of xln(x) so rearrange the antid you a trying to get in terms of intergal(xln(x))
so 2intergral(ln(x))=x²ln(x)-intergral(x) and go from there, then sub in terminals

[brightsky]

integration by recognition
a) d/dx (x^2 ln(x)) = x + 2xln(x)
b) since we have the result from a), we know that x^2 ln(x) = int (x) dx + 2 int(xln(x)) dx by integrating both sides,
so int(x ln(x)) dx = 1/2(x^2 ln(x) - x^2/2)
since for the required region, we are bounded on the left by 1 and on the right by 3:
then the shaded region is:
1/2(3^2 ln(3) - (3)^2/2) - 1/2(1 ln(1) - 1/2)
=9/2 ln(3) - 9/4 + 1/4
=9/2 ln(3) - 2

hopefully no mistakes lol

[jinny1]

so for a you would have got 2x*ln(x)+x
so if you differentiate that you will get x²ln(x)
so you need to know the anitderivative of xln(x) so rearrange the antid you a trying to get in terms of intergal(xln(x))
so 2intergral(ln(x))=x²ln(x)-intergral(x) and go from there, then sub in terminals

you meant integrate right??

integration by recognition
a) d/dx (x^2 ln(x)) = x + 2xln(x)
b) since we have the result from a), we know that x^2 ln(x) = int (x) dx + 2 int(xln(x)) dx by integrating both sides,
so int(x ln(x)) dx = 1/2(x^2 ln(x) - x^2/2)
since for the required region, we are bounded on the left by 1 and on the right by 3:
then the shaded region is:
1/2(3^2 ln(3) - (3)^2/2) - 1/2(1 ln(1) - 1/2)
=9/2 ln(3) - 9/4 + 1/4
=9/2 ln(3) - 2

hopefully no mistakes lol

The point i got stuck in was when i integrated x dx.. i added a contant in and i didnt know what to do with the stupid c! So how come we dont use constants in these cases?

Thanks guys!

[tony3272]

When you take the indefinite integral of something you add the constant c. We still do this when we take the definite integral but then it cancels out, so by convention it's just omitted from the beginning.

For example if you have , then

But if you take the definite integral of 2x, you would get , which expands to , which you can then see that the constant is canceled out.

[jinny1]

Cheers its clearer now. In the current context, x^2ln(x) = (int)2xln(x) + x dx ..

So without first using definitive integral, shouldnt the above equation become simpkified to : x^2ln(x) = ((int)xln(x) dx) + x^2/2 + c

[tony3272]

Yup. And then to work out the area for part b of the question you just rearrange the equation so it's

[jinny1]

Edited:

Nvm. Silly calc error from b4. Thanks!

[jinny1]

Just a quick question   how do i find the (total sum of) function in the ti-nspire. The one with the big E. I need to find whether a function is a prob function by addin it together and see if it is one.

[tony3272]

Press control then multiply. A small table should pop up and you can select it from there.