定积分高数题求解答

1、
y^2=x与x=2y的交点：(0,0)、(4,2)
V=∫(0,2)π(4y^2-y^4)dy
=π(4/3y^3-1/5/y^5)|(0,2)
=π[4/3(2^3-0^3)-1/5(2^5-0^5)]
=π(32/3-32/5)
=32π(5-3)/15
=64π/15
2、
∫(0,π/6)cos^(-3)2θsin2θdθ
=-1/2∫(0,π/6)cos^(-3)2θdcos2θ
=-1/2×1/(-2)cos^(-2)2θ|(0,π/6)
=1/4[cos^(-2)(2π/6)-cos^(-2)0]
=1/4[(1/2)^(-2)-1^(-2)]
=1/4(4-1)
=3/4
∫(√(π/2),√π)x^3cos(x^2)dx
=1/2∫(√(π/2),√π)x^2cos(x^2)dx^2
=1/2∫(√(π/2),√π)x^2dsin(x^2)
=1/2x^2sin(x^2)|[(√(π/2),√π)]-1/2∫(√(π/2),√π)sin(x^2)dx^2
=1/2{(√π)^2sin(√π)^2-[√(π/2)]^2sin[√(π/2)]^2}+1/2cos(x^2)|[√(π/2),√π]
=1/2[πsinπ-(π/2)sin(π/2)]+1/2{cos(√π)^2-cos[√(π/2)]^2}
=1/2(0-π/2)+1/2[cosπ-cos(π/2)]
=-π/4+1/2(-1-0)
=-π/4-1/2
3、
y=∫(sinx,1)√(1+t^2)dt
=-∫(1,sinx)√(1+t^2)dt
y'=√(1+sin^2x)(sinx)'
=cosx√(1+sin^2x)
4、收敛
∫(1,+∞)1/(x^2+x)dx
=∫(1,+∞)[1/x-1/(x+1)]dx
=[lnx-ln(x+1)]|(1,+∞)
=(+∞-ln1)-[+∞-ln(1+1)]
=ln2
5、
S=∫(-1,0)(2^x-3^x)dx+∫(0,1)(3^x-2^x)dx
=(2^xln2-3^xln3)|(-1,0)+(3^xln3-2^xln2)|(0,1)
=2^0ln2-3^0ln3-2^(-1)ln2+3^(-1)ln3+3^1ln3-2^1ln2-3^0ln3+2^0ln2
=ln2-ln3-1/2ln2+1/3ln3+3ln3-2ln2-ln3+ln2
=(1-1/2-2+1)ln2+(-1+1/3+3-1)ln3
=-1/2ln2+4/3ln3