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# C4 integration help watch

1. for part iv where does the root 3 come from?
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2. I presume that root 3 is the area of the graph between y=1 and the x-axis, that you would need to find to determine R.
3. (Original post by Nikita_2313)
I presume that root 3 is the area of the graph between y=1 and the x-axis, that you would need to find to determine R.
When I work that out, I get 8-3root3/6, which is no where near root3
4. (Original post by gooner1010)
When I work that out, I get 8-3root3/6, which is no where near root3
How'd you get that? The line y=1 intersects the graph at so , since

This forms a rectangle with width and height 1. So area .

Where did you go wrong?
5. (Original post by Zacken)
How'd you get that? The line y=1 intersects the graph at so , since

This forms a rectangle with width and height 1. So area .

Where did you go wrong?
Why do you form a rectangle? I thought the area of R was the area of the whole semi circle minus the area of the semi circle below 1. Wouldn't some of the area of the semi circle below 1 be missing?
6. the two equations meet at 1+root3/2 and 1-root3/2. Basically, root 3 is the definite integral of the equation y=1 (when you put in the limits)
integral of 1 is x. so you just subtract the lower x value from the upper one which gives (1 +root3/2) - (1 - root3/2) =root3/2 + root3/2 = root3
hope i explained it alright.
havent done integration in a while because I am on gap year so this was a refresher.
7. (Original post by gooner1010)
Why do you form a rectangle? I thought the area of R was the area of the whole semi circle minus the area of the semi circle below 1. Wouldn't some of the area of the semi circle below 1 be missing?
Eh? You're quite off track. When you integrate between x=1-sqrt(3)/2 and x =1 + sqrt(3)/2, you get the full area of the curve, as shaded here:

To get the required area the question wants, you have to subtract the area of that rectangle I've shaded in. This rectangle has height 1 and width sqrt(3).

Do you understand what it means to integrate between two limits...?
8. (Original post by k_popper)
the two equations meet at 1+root3/2 and 1-root3/2. Basically, root 3 is the definite integral of the equation y=1 (when you put in the limits)
integral of 1 is x. so you just subtract the lower x value from the upper one which gives (1 +root3/2) - (1 - root3/2) =root3/2 + root3/2 = root3
hope i explained it alright.
havent done integration in a while because I am on gap year so this was a refresher.
Oh I get it it, I did it wrong before as I tried working out the area of the whole semi circle. Thank you and Zacken.
9. (Original post by Zacken)
Eh? You're quite off track. When you integrate between x=1-sqrt(3)/2 and x =1 + sqrt(3)/2, you get the full area of the curve, as shaded here:

To get the required area the question wants, you have to subtract the area of that rectangle I've shaded in. This rectangle has height 1 and width sqrt(3).

Do you understand what it means to integrate between two limits...?
Thanks, I get it now.
10. (Original post by gooner1010)
Oh I get it it, I did it wrong before as I tried working out the area of the whole semi circle. Thank you and Zacken.
my pleasure. I actually enjoyed solving this T_T i wish i could take a small maths course on the side at uni lmao
11. (Original post by gooner1010)
Thanks, I get it now.
No worries

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