求下列定积分 1)∫[1,3]x^3dx 2)∫[1,4]√(x)dx 3)∫[π,2π]sinxdx 4)∫[0,1](1/4t^2-9)dt……求下列定积分1)∫[1,3]x^3dx2)∫[1,4]√(x)dx3)∫[π,2π]sinxdx4)∫[0,1](1/4t^2-9)dt5)∫[-1,0]e^(-x)dx6)∫[-1,-2](x/x+3)dx

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求下列定积分 1)∫[1,3]x^3dx 2)∫[1,4]√(x)dx 3)∫[π,2π]sinxdx 4)∫[0,1](1/4t^2-9)dt……求下列定积分1)∫[1,3]x^3dx2)∫[1,4]√(x)dx3)∫[π,2π]sinxdx4)∫[0,1](1/4t^2-9)dt5)∫[-1,0]e^(-x)dx6)∫[-1,-2](x/x+3)dx

求下列定积分 1)∫[1,3]x^3dx 2)∫[1,4]√(x)dx 3)∫[π,2π]sinxdx 4)∫[0,1](1/4t^2-9)dt……求下列定积分1)∫[1,3]x^3dx2)∫[1,4]√(x)dx3)∫[π,2π]sinxdx4)∫[0,1](1/4t^2-9)dt5)∫[-1,0]e^(-x)dx6)∫[-1,-2](x/x+3)dx
求下列定积分 1)∫[1,3]x^3dx 2)∫[1,4]√(x)dx 3)∫[π,2π]sinxdx 4)∫[0,1](1/4t^2-9)dt……
求下列定积分
1)∫[1,3]x^3dx
2)∫[1,4]√(x)dx
3)∫[π,2π]sinxdx
4)∫[0,1](1/4t^2-9)dt
5)∫[-1,0]e^(-x)dx
6)∫[-1,-2](x/x+3)dx

求下列定积分 1)∫[1,3]x^3dx 2)∫[1,4]√(x)dx 3)∫[π,2π]sinxdx 4)∫[0,1](1/4t^2-9)dt……求下列定积分1)∫[1,3]x^3dx2)∫[1,4]√(x)dx3)∫[π,2π]sinxdx4)∫[0,1](1/4t^2-9)dt5)∫[-1,0]e^(-x)dx6)∫[-1,-2](x/x+3)dx
1)∫[1,3]x^3dx
=1/4*x^4=1/4*(3^4-1^4)=20
2)∫[1,4]√(x)dx
=2/3*x^(3/2)=2/3*4^(3/2)-2/3*1^(3/2)=14/3
3)∫[π,2π]sinxdx
=-cosx=-(cos2π-cosπ)=-2
4)∫[0,1](1/4t^2-9)dt
=1/12t^3-9t=1/12-9=-107/12
5)∫[-1,0]e^(-x)dx
=-e^(-x)=-(1-e)=e-1
6)∫[-1,-2](x/x+3)dx
=∫[-1,-2][1-3/(x+3)]dx
=x-3ln(x+3)
=-2-3ln1+1+3ln2
=3ln2-1

解:
1) ∫[1,3]x^3dx = 1/4*x^4 | [1,3] = 20
2) ∫[1,4]√(x)dx = 2/3* x√x | [1,4] = 14/3
3) ∫[π,2π]sinxdx = -cosx | [π,2π] = -2
4) ∫[0,1](1/4t^2-9)dt = 1/12 *t^3 -9t...

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解:
1) ∫[1,3]x^3dx = 1/4*x^4 | [1,3] = 20
2) ∫[1,4]√(x)dx = 2/3* x√x | [1,4] = 14/3
3) ∫[π,2π]sinxdx = -cosx | [π,2π] = -2
4) ∫[0,1](1/4t^2-9)dt = 1/12 *t^3 -9t | [0,1] = -107/12
5) ∫[-1,0]e^(-x)dx =-e^(-x) | [-1,0] = e-1
6) ∫[-1,-2](x/x+3)dx = ∫[-1,-2][(1-3/(x+3)]dx
=x -3ln(x+3)|[-1,-2] = 3ln2 -1
以上答案供参考,祝你学习进步

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1)∫[1,3]x^3dx
=(1/4)x^4 |[1,3]=81/4-1/4=20
2)∫[1,4]√(x)dx
=(2/3)√(x^3) |[1,4] =16/3-2/3=14/3
3)∫[π,2π]sinxdx
=-cosx |[π,2π]=-1-(1)=-2
4)∫[0,1](dt/(4t^2-9)=(1/12)∫[0,1](2t+3)-...

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1)∫[1,3]x^3dx
=(1/4)x^4 |[1,3]=81/4-1/4=20
2)∫[1,4]√(x)dx
=(2/3)√(x^3) |[1,4] =16/3-2/3=14/3
3)∫[π,2π]sinxdx
=-cosx |[π,2π]=-1-(1)=-2
4)∫[0,1](dt/(4t^2-9)=(1/12)∫[0,1](2t+3)-(2t-3)d2t/[(2t+3)(2t-3)]
=(1/12)ln|(2t-3)/(2t+3)| |[0,1] =(1/12)ln(1/5)=(-1/12)ln5
5)∫[-1,0]e^(-x)dx
= -e^(-x) |[-1,0] =-1-(-e)=e-1
6)∫[-1,-2] xdx/(x+3)=∫[-1,-2][(x+3)-3]dx/(x+3)
=x-3ln|x+3| |[-1,-2] =-2-(-1-3ln2)=3ln2-1

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