Question

求定积分题

1、∫(1,0) xe^(x^2)dx
2、∫(π/2,0) sinxcos^2 xdx
定积分，求过程

Answer 1

1、∫(1,0) xe^(x^2)dx=∫(1,0)(1/2)e^(x²) dx²=∫(1,0)(1/2)d(e^(x²))=(1/2)e^(x²)|(1,0)=(1/2)(e-1)=
(e-1)/2.
2.∫(π/2,0)sinxcos^2xdx=∫(π/2,0)-cos^2xdcosx=∫(π/2,0)-(1/3)d(cos^3x)=(-1/3)cos^3x|(π/2,0)=(-1/3)(0-1)=1/3

Answer 2

1，原式得 积分（0到1）e＾（x＾2）dx＾2＝e＾1＾2－e＾0＝e－1
2，原式得
积分（0到pi／2）－cos＾2xdcosx＝－（cos＾3（pi／2）－ cos＾3（0））／3＝1／3

Answer 3

∫(1,0) xe^(x^2)dx
=（1/2）∫(1,0) 2x e^(x^2) dx
=（1/2）∫(1,0) e^(x^2) d(x^2)
=（1/2）e^(x^2) I (1,0)
=（1/2）（e - 1）

∫(π/2,0) sinxcos^2 xdx
=-∫[0,π/2] cos^2 xdcosx
=-1/3cos^3 I [0,π/2]
=1/3

Answer 4

∫[0,1] xe^(x^2)dx
=1/2∫(1,0) e^(x^2)dx^2
=1/2e^(x^2)[0,1]
=(e-1)/2
∫[0,π/2] sinxcos^2 xdx
=-∫[0,π/2] cos^2 xdcosx
=-1/3cos^3x[0,π/2]
=1/3

Answer 5

1.分步积分：∫xe^(x^2)dx =e^(x^2)/2
所以∫(1,0) xe^(x^2)dx
下限是0吧
=e/2 - 1/2
2.= 0.3271