求下列各积分
3个回答
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2、∫e^xcose^xdx
=∫cose^xde^x
=sine^x+C
3、∫(4,7)x/√(x-3)dx
=2∫(4,7)xd√(x-3)
=2[x√(x-3)]|(4,7)-2∫(4,7)√(x-3)dx
=2[7√(7-3)-4√(4-3)]-2/(3/2)(x-3)^(3/2)|(4,7)
=2[7×2-4×1]-4/3[(7-3)^(3/2)-(4-3)^(3/2)]
=20-4/3[8-1]
=20-28/3
=32/3
4、∫(0,π/2)xcosxdx
=∫(0,π/2)xdsinx
=xsinx|(0,π/2)-∫(0,π/2)sinxdx
=(π/2)sin(π/2)-0sin0+cosx|(0,π/2)
=π/2+cos(π/2)-cos0
=π/2+0-1
=π/2-1
5、∫(2,+∞)1/x^4dx
=-1/3x^(-3)|(2,+∞)
=-1/3[∞^(-3)-2^(-3)]
=-1/3[0-1/8]
=1/24
=∫cose^xde^x
=sine^x+C
3、∫(4,7)x/√(x-3)dx
=2∫(4,7)xd√(x-3)
=2[x√(x-3)]|(4,7)-2∫(4,7)√(x-3)dx
=2[7√(7-3)-4√(4-3)]-2/(3/2)(x-3)^(3/2)|(4,7)
=2[7×2-4×1]-4/3[(7-3)^(3/2)-(4-3)^(3/2)]
=20-4/3[8-1]
=20-28/3
=32/3
4、∫(0,π/2)xcosxdx
=∫(0,π/2)xdsinx
=xsinx|(0,π/2)-∫(0,π/2)sinxdx
=(π/2)sin(π/2)-0sin0+cosx|(0,π/2)
=π/2+cos(π/2)-cos0
=π/2+0-1
=π/2-1
5、∫(2,+∞)1/x^4dx
=-1/3x^(-3)|(2,+∞)
=-1/3[∞^(-3)-2^(-3)]
=-1/3[0-1/8]
=1/24
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(2)
∫e^x .cos(e^x)dx
=∫dsin(e^x)
=sin(e^x) +C
(3)
∫(4->7) x/√(x-3) dx
=2∫(4->7) x d√(x-3)
=2[ x.√(x-3)]|(4->7) - 2∫(4->7) √(x-3) dx
=2(14-4) - (4/3)[(x-3)^(3/2)]|(4->7)
=20 - (4/3)( 8-1)
=20 - 28/3
=32/3
(5)
∫(2->+∞) dx/x^4
= -(1/3) [ 1/x^3] |(2->+∞)
= 1/24
(4)
∫(0->π/2) xcosx dx
=∫(0->π/2) xdsinx
=[x.sinx]|(0->π/2) -∫(0->π/2) sinx dx
=π/2 + [cosx]|(0->π/2)
=π/2 - 1
∫e^x .cos(e^x)dx
=∫dsin(e^x)
=sin(e^x) +C
(3)
∫(4->7) x/√(x-3) dx
=2∫(4->7) x d√(x-3)
=2[ x.√(x-3)]|(4->7) - 2∫(4->7) √(x-3) dx
=2(14-4) - (4/3)[(x-3)^(3/2)]|(4->7)
=20 - (4/3)( 8-1)
=20 - 28/3
=32/3
(5)
∫(2->+∞) dx/x^4
= -(1/3) [ 1/x^3] |(2->+∞)
= 1/24
(4)
∫(0->π/2) xcosx dx
=∫(0->π/2) xdsinx
=[x.sinx]|(0->π/2) -∫(0->π/2) sinx dx
=π/2 + [cosx]|(0->π/2)
=π/2 - 1
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2.=∫cos(e^x)de^x=sin(e^x)+C
3.=∫(1到2)(t²+3)/td(t²+3)
=∫2t³+3tdt
=t^4/2+3t²/2
=15/2+9/2
=12
4.=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx=π/2-1
3.=∫(1到2)(t²+3)/td(t²+3)
=∫2t³+3tdt
=t^4/2+3t²/2
=15/2+9/2
=12
4.=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx=π/2-1
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