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INTEGRAL HELP smhmyhead
imgur.com/rSNDPF9 doesnt look so hard but can't solve it ffs help plz
2019-02-21 19:48
#1 MBT_yt
2+2=5
2019-02-21 19:49
#2 2k19
2+2 is 7. damn
2019-02-21 19:50
#3 smarthltvuser
A
2019-02-21 19:50
#9 smhmyhead
nasil yaptin soyle LUTFEEEEEEN A dogru
2019-02-21 19:55
#14 smarthltvuser
thats easy wtf
2019-02-21 19:58
#4 2k19
it's A
2019-02-21 19:51
#5 omaiguudness
Its D) Thank me l8r
2019-02-21 19:51
#6 smhmyhead
but how did you solve
2019-02-21 19:52
#7 fazecrap
zzzzzz 10th grade u 12yr old
2019-02-21 19:52
#8 smhmyhead
im 11th grade and trying to study home alone, 3 months left for upcoming exam plz help
2019-02-21 19:54
#10 moetxxx
it is e ecuation of first circle equation - second
2019-02-21 19:59
#11 smhmyhead
i know those but still can't solve this question
2019-02-21 19:58
#13 moetxxx
sorry i thought you have 5 integral to solves
2019-02-21 19:58
#12 wtfuninstall
C you're looking at the negative because its all below x axis. look at negative of big circle which is integral from -1 to 1 of -(sqrt(9-x^2)) (call this g) and then subtracting the semi-circle which is -sqrt((1-x^2)) (call this f) so: g-f to find area and you get sqrt(1-x^2) - (sqrt(9-x^2) and you can workout formula of circle easily by x^2+y^2 = r^2 where r is radius. you can rearrange to find y by -x^2 and taking + and - sqrt. discard the + sqrt because you're only looking below the x-axis
2019-02-21 19:59
#19 smhmyhead
sorry for bothering again but how did you find sqrt(9-x^2) and sqrt(1-x^2)? i mean i'm not so bad at graphs but didn't get this one
2019-02-21 20:06
#22 wtfuninstall
radius of big circle = 3. formula for circle = (x-a)^2+(y-b)^2=r^2. since the origin of both circles is at 0,0 a and b= 0. so the formula for the big circle is x^2+y^2=3^2 and the formula for the small circle is x^2+y^2=1^2. find y by rearranging using basic algebra. you get for the big circle y^2=3^2-x^2 = 9-x^2. to find y you just take the square root and get sqrt(q-x^2). same process to work out the second circle. it might be A. could've messed up with the negatives somewhere
2019-02-21 20:10
#24 smhmyhead
thanks for your effort
2019-02-21 20:10
#26 wtfuninstall
i think i understand why the answer is A. they're asking for the area so you can't get a negative area. you have to take the negative of my answer to get a positive area and from that you get answer A.
2019-02-21 20:15
#20 smhmyhead
the answer is A by the way.
2019-02-21 20:07
#21 smhmyhead
oh this is coming from analytic geometry? i'm so lazy to learn that shit and probably will skip that lesson.
2019-02-21 20:08
#15 <3 Zetas
B
2019-02-21 19:58
#16 2k19
S =sin(x+a)ex(xdx)
2019-02-21 19:58
#17 Magma_44
C You want to subtract the two circles to find the area between. Since you're taking the bottom part of the integral, you want to take the negative parts of the square roots. So you'd take the integral from -1 to 1 of -sqrt(9-x^2) - (-sqrt(1-x^2)), which if you simplify out comes out to C.
2019-02-21 19:59
#18 smatchcube
Not C as the result is negative.
2019-02-21 20:02
#25 Magma_44
it's asking for the integral, not the area
2019-02-21 20:12
#27 smatchcube
Well, maybe you are right, in my country we often use the letter S for Surface and A for Aire (area).
2019-02-21 20:17
#28 DvD--
yeah but your country is a fucking shithole
2019-02-21 20:19
#29 smhmyhead
wtf respect france?
2019-02-21 20:21
#30 smatchcube
Can you remind me how many players from Netherlands are present at the major?
2019-02-21 20:24
#32 DvD--
yeah but your country is a fucking shithole
2019-02-21 21:47
#23 pos_anec
ez :) The bigger circle equation: c1: x^2 + y^2 = 3^2 The smaller circle equation: c2: x^2 + y^2 = 1^2 Now, just express the y from both: f1: y = abs(sqrt(3^2 - x^2)) f2: y = abs(sqrt(1^2 - x^2)) Note the absolute value, which accounts for the fact that you can either count the upper half or the bottom half of the circle. Since you are counting the bottom (negative) part (see here > revisionmaths.com/advanced-level-maths-r..) both expressions would turn to their negative versions: f1: y = - sqrt(3^2 - x^2) f2: y = - sqrt(1^2 - x^2) And last part, just take [bigger circle area] - [smaller circle area] = integral from -1 to 1 [ f1(x) - f2(x) ] = integral from -1 to 1 [ - sqrt(3^2 - x^2) - (- sqrt(1^2 - x^2)) ] = integral from -1 to 1 [ sqrt(1^2 - x^2) - sqrt(3^2 - x^2) ] = integral from -1 to 1 [ sqrt(1 - x^2) - sqrt(9 - x^2) ] Which is the A option :0)
2019-02-21 20:10
#31 pos_anec
Obviously it's the C option :D But as mentioned above, the area could only be positive, therefore it is the A option probably.
2019-02-21 20:28
#33 HippoCS