幫解幾題微積分!!!
Find the area of the regions enclosed by the line and curves.
1.y=x(a^2-x^2)^1/2,a>0,y=0
2.y=|x^2-4| and y=x^2/2+4
3.y=cosx and y=sec^2x, -π/3 ≤x≤π/3
解答
1.(2/3)(a^3)
2.21(1/3)
3.6√3
請大大給我詳細的算式 有面積圖形最好!!! 答案如果有誤請告知!!! 我算不出這三個答案
所以第二題跟第三題答案是錯的??
2 個解答
- 阿銘Lv 78 年前最佳解答
則A=2*Sx(a^2-x^2)1/2dxlx=0至a
Find the area of the regions enclosed by the line and curves.
1.y=x(a^2-x^2)^1/2,a>0,y=0
y=0
x=0 or x=+,-a
且因y=x(a^2-x^2)^1/2,當x以-x代入得y(-x)=-y(x)
故圖形對y軸為反對稱
令y=x(a^2-x^2)^1/2,a>0,y=0間之面積為A
令a^2-x^2=t,則 -2xdx=dt代得
A=2*S-1/2*t^1/2dtlt=a^2至0
=-1*2/3*t^3/2lt=a^2至0
=-2/3*(0-a^3)
=(2/3)a^3
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2.y=|x^2-4| and y=x^2/2+4
a.x=>2,y=x^2-4and y=x^2/2+4,且X^2+4>X^2-4 求解 x=+,-4,-4不合 故解為x=4
b.2=>x=>-2,,y=4-x^2 andy=x^2/2+4,x=0,故解為x=0
cx<=-2,y=x^2-4,andy=x^2/2+4,x++4,-4,4不合,故解為x=-4
令所求面積為A,因以X=-X代入得X^2=(-X)^2 故圖形對Y軸為對稱
則A=2*S[(X^2/2+4)-(X^2-4)]dxlX=0至4
=2*S8-X^2/2dxlx=0至4
=2*[8x-x^3/6]lx=0至4
=2*[(32-32/3)-(0)]
=2*64/3=128/3
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3.y=cosx and y=sec^2x, -π/3 ≤x≤π/3
-π/3 ≤x≤π/3
y=cosx and y=sec^2x cosx=sec^2x,cos^3x=1,cosx=1,x=0
且-1<=cosx<=1 secx=1/cosx,故 sec^x=>cosx
cosx=cos(-x),sec^2x=sec^2(-x),故圖形對y軸對稱
令所求面積為A
則A=2*Ssec^2x-cosxdxlx=0至π/3
=2*(tanx-sinx)lx=0至π/3
=2*(tanπ/3-sinπ/3)-0
=2*(√3-/√3/2)=√3
2013-08-25 17:53:00 補充：
第2題漏個2次方 修正如下:且-1<=cosx<=1 secx=1/cosx,故 sec^2x=>cosx
2013-08-26 08:50:30 補充：
抱歉第2題有誤 未分段積分 更正如下:
令所求面積為A,因以X=-X代入得X^2=(-X)^2 故圖形對Y軸為對稱
則A=2*{S(X^2/2+4)-lX^2-4l]dxlX=0至2+S[(x^2/2+4)-lx^2-4llx=2至4]}
=2*{S(x^2/2+4+x^2-4)dxlx=0至2+S{x^2/2+4-x^2+4}dxlx=2至4l}
=2*[S3X^2/2dxlX=0至2]+[S-x^2/2+8dxlx=2至4]}
2013-08-26 08:50:36 補充：
=2*{(x^3/2lx=0至2)+(-x^3/6+8xlx=2至4)}
=2*[(8/2)+(-64/6+32+8/6-16)]
=2*(20-56/6)=2*64/6=128/6=21+(2/6=21+1/3
