question
let g(x)=cos x^2 , f(x)=x^1/2 and a and b (a<b) be the root of the quadratic equation 18x^2-9(pi)x+(pi)^2=0 then the area bounded by the curve y=(gof)(x)and the lines x=a , x=b and y=0 is.
let g(x)=cos x^2 , f(x)=x^1/2 and a and b (a<b) be the root of the quadratic equation 18x^2-9(pi)x+(pi)^2=0 then the area bounded by the curve y=(gof)(x)and the lines x=a , x=b and y=0 is.
given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is 圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872668990.jpg
 
首先請先 複習下列公式
 
設扇形的中心角θ弳，半徑為r，弧長為s，面積為a，則
s=rθ；a= r2θ= rs
sol: 設: ∠aob=θ 弳 (弧度) od=x
(x+2.5) θ =6 ……………………………..(1)
求黃色部分面積= (x+5)2 θ - x2θ
= θ ( x2+10x+25-x2 )
= θ (10x+25)
=5θ( x + 2.5)=30



圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669011.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669012.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669033.jpg
 



 2010-06-04 23:25:06 補充：
y= |cosx|

恕不再多畫 將以文字說明

1. 當cosx >0 時 y=cosx
2. 當cosx <0 時 y = -cosx

 2010-06-04 23:26:21 補充：
注意

y=3sin2(x-π)+2

之 a , b 步驟為同一圖形

 2010-06-05 13:06:49 補充：
奇怪 1/2 通通不見了

更正

s=rθ；a=0.5 r2θ=0.5 rs

求黃色部分面積=0.5(x+5)2 θ - 0.5x2 θ
=0.5θ ( x2+10x+25-x2 )
=θ (10x+25)
=5θ( x + 2.5)=30

p.s. x2 即為 x平方

 2010-08-28 11:35:07 補充：
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rid of without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. 18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer