Question

计算定积分。

定积分符号不好打，我就用区间直接表示了。
1.函数(4-2x)(4-x^2)dx , [0,2]
2.(x^2-2x-3)/x dx [1,2]
3.(√x+1/√x)^2dx [2,3]
4.√x（1-√x）dx [1,4]
5.(3x+sinx)dx [0,2π]
6.(e^x-2/x)dx [1,2]

1.e^2xdx [0,1]
2.cos2xdx [π/6,π/4]
3.2^xdx [1,3]
就这些了，小弟刚接触定积分，不是很熟练，麻烦GGJJ帮帮忙...

Answer 1

Answer 2

∫(0~2) (4 - 2x)(4 - x^2) dx
= ∫(0~2) (2x^3 - 4x^2 - 8x + 16) dx
= 2 * x^4/4 - 4 * x^3/3 - 8 * x^2/2 + 16x |(0~2)
= (1/2) * 2^4 - (4/3) * 2^3 - 4 * 2^2 + 16 * 2
= 40/3
∫(1~2) (x^2 - 2x - 3)/x dx
= ∫(1~2) (x - 2 - 3/x) dx
= x^2/2 - 2x - 3lnx |(1~2)
= [1/2 * 2^2 - 2(2) - 3ln2] - [1/2 - 2 - 0]
= - 1/2 - 3ln2
∫(2~3) (√x + 1/√x)^2 dx
= ∫(2~3) (x + 2 + 1/x) dx
= x^2/2 + 2x + lnx |(2~3)
= [1/2 * 3^2 + 2(3) + ln3] - [1/2 * 2^2 + 2(2) + ln2]
= 9/2 + ln(3/2)
∫(1~4) √x * (1 - √x) dx
= ∫(1~4) (√x - x) dx
= (2/3)x^(3/2) - x^2/2 |(1~4)
= [2/3 * 4^(3/2) - 1/2 * 4^2] - [2/3 - 1/2]
= - 17/6
∫(0~2π) (3x + sinx) dx
= 3 * x^2/2 - cosx |(0~2π)
= [3/2 * (2π)^2 - cos(2π)] - [0 - cos0]
= 6π^2
∫(1~2) (e^x - 2/x) dx
= e^x - 2lnx |(1~2)
= [e^2 - 2ln2] - [e - 0]
= e^2 - e - 2ln2

∫(0~1) e^(2x) dx
= ∫(0~1) e^(2x) d(2x)/2
= 1/2 * e^(2x) |(0~1)
= [1/2 * e^2] - [1/2 * 1]
= (e^2 - 1)/2
∫(π/6~π/4) cos2x dx
= (π/6~π/4) cos2x d(2x)/2
= 1/2 * sin2x |(π/6~π/4)
= [1/2 * sin(2 * π/4)] - [1/2 * sin(2 * π/6)]
= (2 - √3)/4
∫(1~3) 2^x dx
= 2^x/ln2 |(1~3)
= [1/ln2 * 2^3] - [1/ln2 * 2]
= 6/ln2
定积分不难，先求出不定积分，然后再把代入上限的原函数，减去代入下限的原函数就可以了。

Answer 3

求定积分;
1.[0，2]∫(4-2x)(4-x²)dx =.[0，2]∫(2x³-4x²-8x+16)dx=[x⁴/2-(4/3)x³-4x²+16x]︱[0，2]
=8-(32/3)-16+32=24-(32/3)=40/3
2.[1，2]∫(x²-2x-3)/x dx=[1，2]∫[x-2-(3/x)]dx=[(x²/2)-2x-3lnx︱[1，2]
=(2-4-3ln2)-(1/2-2)=-(1/2)-3ln2
3.[2，3]∫(√x+1/√x)²dx=[2，3]∫[x+2+(1/x)]dx=[(x²/2)+2x+lnx]︱[2，3]
=(9/2)+6+ln3-2-4-ln2=(9/2)+ln(3/2)
4.[1，4]∫(√x)(1-√x)dx=[1，4]∫[(√x)-x]dx=[(2/3)x^(3/2)-x²/2]︱[1，4]
=16/3-8-(2/3-1/2)=23/6
5.[0，2π]∫(3x+sinx)dx=[(3x²/2)-cosx]︱[0，2π]=6π²-1-(-1)=6π²
6.[1，2]∫(e^x-2/x)dx=(e^x-2lnx)︱[1，2]=e²-2ln2-e=e²-e-2ln2
7.[0，1]∫e^(2x)dx=.[0，1](1/2)∫e^(2x)d(2x)=(1/2)e^(2x)︱[0，1]=(1/2)(e²-1)
8.[π/6，π/4]∫cos2xdx=[π/6，π/4](1/2)∫cos2xd(2x)=(sin2x)︱[π/6，π/4]=1-(√3)/2
9.[1，3]∫2^xdx=[(2^x)ln2]︱[1，3]=8ln2-2ln2=6ln2.