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∫[0,a]√(a^2-x^2)dx=a^2∫[0,a]√[1-(x/a)^2]d(x/a)
x/a=sinu x=a,u=π/2,x=0,u=0
=a^2∫[0,π/2]cosudsinu
=a^2∫[0,π/2](1+cos2u)du/2
=a^2π/4
y=∫[0,x](t-1)(t-2)dt=∫[0,x] (t^2-3t+2)dt=x^3/3-3x^2/2+2x
y'=(x-1)(x-2),x=1,y极值=7/3-3/2=5/6
x=2 y极值=8/3-6+4=2/3
∫[0,y]e^tdt+∫[0,x] costdt=0
e^y+sinx=0
e^y=-sinx
y=ln|-sinx| y'=-cosx/-sinx=cotx
∫dx/(1+√x)=∫2√xd(√x)/(1+√x)=2-∫d√x/(1+√x)=2-ln(1+√x)
∫xarctanxdx=(x^2/2)arctanx-(1/2)∫x^2dx/(1+x^2)=(x^2/2)arctanx-1/2+(1/2)∫dx/(1+x^2)
=(x^2/2-1/2)arctanx-1/2+C
∫[0,1]xarctanxdx=-1/2
x/a=sinu x=a,u=π/2,x=0,u=0
=a^2∫[0,π/2]cosudsinu
=a^2∫[0,π/2](1+cos2u)du/2
=a^2π/4
y=∫[0,x](t-1)(t-2)dt=∫[0,x] (t^2-3t+2)dt=x^3/3-3x^2/2+2x
y'=(x-1)(x-2),x=1,y极值=7/3-3/2=5/6
x=2 y极值=8/3-6+4=2/3
∫[0,y]e^tdt+∫[0,x] costdt=0
e^y+sinx=0
e^y=-sinx
y=ln|-sinx| y'=-cosx/-sinx=cotx
∫dx/(1+√x)=∫2√xd(√x)/(1+√x)=2-∫d√x/(1+√x)=2-ln(1+√x)
∫xarctanxdx=(x^2/2)arctanx-(1/2)∫x^2dx/(1+x^2)=(x^2/2)arctanx-1/2+(1/2)∫dx/(1+x^2)
=(x^2/2-1/2)arctanx-1/2+C
∫[0,1]xarctanxdx=-1/2
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解答:
(1)原不定积分=∫a²√[1-(x/a)²]d(x/a)=a²arcsin(x/a),∴原式=πa²/2;
(2)y=x³/3-3x²/2+2x,y'=(x-1)(x-2),易知y增区间为x≥2或x≤1,减区间为(1,2),
因此,当x=1时,函数取得极大值5/6,当x=2时取得极小值2/3;
(3)原式化简得e^y-1+sinx=0,设F=e^y-1+sinx,则隐函数y的一阶导数= - Fx/Fy= - cosx/(e^y);
(4)设√x=t,则dx=2tdt,原式=∫ 2t/(t+1)dt=∫ [2-2/(t+1)]dt=2t-2ln(t+1)=2√x-2ln(√x+1),
定积分为2-2ln(3/2);
(5)利用分部积分,原式=1/2 ∫ arctanxd(x²)= 1/2[(x²+1)arctanx-1],定积分=(π-2)/4;
(6)设x²=tant,则dx=sec²tdt,原式=∫ costdt=sint=sin(tanx),定积分=sin(tan1);
(7)原式=∫ 1/(lnx)²d(lnx)=-(1/lnx),则定积分=e
(1)原不定积分=∫a²√[1-(x/a)²]d(x/a)=a²arcsin(x/a),∴原式=πa²/2;
(2)y=x³/3-3x²/2+2x,y'=(x-1)(x-2),易知y增区间为x≥2或x≤1,减区间为(1,2),
因此,当x=1时,函数取得极大值5/6,当x=2时取得极小值2/3;
(3)原式化简得e^y-1+sinx=0,设F=e^y-1+sinx,则隐函数y的一阶导数= - Fx/Fy= - cosx/(e^y);
(4)设√x=t,则dx=2tdt,原式=∫ 2t/(t+1)dt=∫ [2-2/(t+1)]dt=2t-2ln(t+1)=2√x-2ln(√x+1),
定积分为2-2ln(3/2);
(5)利用分部积分,原式=1/2 ∫ arctanxd(x²)= 1/2[(x²+1)arctanx-1],定积分=(π-2)/4;
(6)设x²=tant,则dx=sec²tdt,原式=∫ costdt=sint=sin(tanx),定积分=sin(tan1);
(7)原式=∫ 1/(lnx)²d(lnx)=-(1/lnx),则定积分=e
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