Simultaneous Integration?!

http://www.ocr.org.uk/Images/78547-question-paper-unit-4722-core-mathematics-2.pdf
Question 7b:

Now before you say it, I know that the mark scheme is on it the website. But I don't think it'll help me if something like question 7b comes up when I'm doing my C2 test.

I have no clue on how to do these sort of questions, I tried to make y=0 for both and I got 0=8/x^2 - 2 to be x= square root of 6
And 0=6x^3/2 to 6 root 6x.
No way near, I know. Help!
I hate typing out my calculations like that, it looks like a mess typed out

Reply 1

davros

16

Original post by Fruitbasket786

http://www.ocr.org.uk/Images/78547-question-paper-unit-4722-core-mathematics-2.pdf
Question 7b:

Now before you say it, I know that the mark scheme is on it the website. But I don't think it'll help me if something like question 7b comes up when I'm doing my C2 test.

I have no clue on how to do these sort of questions, I tried to make y=0 for both and I got 0=8/x^2 - 2 to be x= square root of 6
And 0=6x^3/2 to 6 root 6x.
No way near, I know. Help!
I hate typing out my calculations like that, it looks like a mess typed out

There isn't really any subtlety here.

You're given the intersection occurs at x = 1, so you integrate the 1st curve from 0 to 1, and the 2nd curve from 1 to 2 (because y= 0 at x = 2). Not sure where you've got root(6) from!

Reply 2

username1560589

20

Original post by Fruitbasket786

http://www.ocr.org.uk/Images/78547-question-paper-unit-4722-core-mathematics-2.pdf
Question 7b:

Now before you say it, I know that the mark scheme is on it the website. But I don't think it'll help me if something like question 7b comes up when I'm doing my C2 test.

I have no clue on how to do these sort of questions, I tried to make y=0 for both and I got 0=8/x^2 - 2 to be x= square root of 6
And 0=6x^3/2 to 6 root 6x.
No way near, I know. Help!
I hate typing out my calculations like that, it looks like a mess typed out

Think about it in terms of areas. The total shaded area is equal to the area under the lowest curve on the left plus the area under the lowest curve on the right. The curves switch at the intersection, x = 1.

A_T = A_L + A_R

The lower curve on the left is

y = 6x^{\frac{3}{2}}

, you can check this by substituting values or thinking about how the curves change as

x\rightarrow 0

.

\therefore A_L = \displaystyle\int_0^1 6x^{\frac{3}{2}}dx

A similar argument can be made about the right area. (curve is

y = \frac{8}{x^2} - 2

). The curve intercepts the x axis at x = 2, so the area stops at x = 2.

\therefore A_R = \displaystyle\int_1^2 (8x^{-2} - 2)dx

A_T = \displaystyle \int_0^1 6x^{\frac{3}{2}}dx + \displaystyle\int_1^2 (8x^{-2} - 2)dx

And onwards from there.

Simultaneous Integration?!

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