定积分计算若f(x)=x^2+2(∫fxdx从0到1积分),求∫fxdx(从0到1积分的值.)答案为-1/3.麻烦给出过程,两边同时积分看不懂啊.

Question

定积分计算
若f(x)=x^2+2(∫fxdx从0到1积分),求∫fxdx(从0到1积分的值.)答案为-1/3. 麻烦给出过程,两边同时积分看不懂啊.

Answer 1

∫(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.