Question

用定积分的分部积分法做下题

求过程

Answer 1

∫[0,π/2]e^(2x)cosxdx
=∫[0,π/2]e^(2x)d(sinx)
=e^(2x)sinx |[0,π/2] - 2∫[0,π/2]e^(2x)sinxdx
=e^π -0 +2∫[0,π/2]e^(2x)d(cosx)
=e^π + 2e^(2x)cosx |[0,π/2] - 4∫[0,π/2]e^(2x)cosxdx
=e^π + 0 - 2 -4∫[0,π/2]e^(2x)cosxdx
故∫[0,π/2]e^(2x)cosxdx=(e^π -2)/5

Answer 2

(1)：
∫(0→π) xsinx dx
= ∫(0→π) x d(- cosx)
= - xcosx：[0→π] + ∫(0→π) cosx dx
= - π(- 1) + sinx：[0→π]
= π
(2)：
∫(0→1) xe^x dx
= ∫(0→1) x d(e^x)
= xe^x：[0→1] - ∫(0→1) e^x dx
= e - e^x：(0→1)
= e - (e - 1)
= 1
(3)：
∫(1→e) x(x - 1)lnx dx
= ∫(1→e) (x^2 - x)lnx dx
= ∫(1→e) lnx d(x^3/3 - x^2/2)
= (x^3/3 - x^2/2)lnx：(1→e) - ∫(1→e) (x^3/3 - x^2/2)(1/x) dx
= (1/3)e^3 - (1/2)e^2 - ∫(1→e) (x^2/3 - x/2) dx
= (1/3)e^3 - (1/2)e^2 - (x^3/9 - x^2/4)：(1→e)
= (1/3)e^3 - (1/2)e^2 - [(e^3/9 - e^2/4) - (1/9 - 1/4)]
= (2/9)e^3 - (1/4)e^2 - 5/36
(4)：
∫(0→1) x^2e^(2x) dx
= (1/2)∫(0→1) x^2 d(e^(2x))
= (1/2)x^2e^(2x)：(0→1) - (1/2)∫(0→1) 2xe^(2x) dx
= (1/2)e^2 - (1/2)∫(0→1) x d(e^(2x))
= (1/2)e^2 - (1/2)xe^(2x)：(0→1) + (1/2)∫(0→1) e^(2x) dx
= (1/2)e^2 - (1/2)e^2 + (1/4)e^(2x)：(0→1)
= (1/4)(e^2 - 1)