∫(1→4)x(√x-1)^4dx.我算到536/35,可答案是351/140……求高手解题!还有∫(-1→3)|3x+1|dx=10/3∫(-3→3)(x^3-x^2-2x+1)dx=24都是怎么算出来的？

Question

∫(1→4)x(√x-1)^4dx.我算到536/35,可答案是351/140……求高手解题!
还有∫(-1→3)|3x+1|dx=10/3
∫(-3→3)(x^3-x^2-2x+1)dx=24都是怎么算出来的？

Answer 1

1
(√x-1)^4=(x+1-2√x)^2=x^2+1+4x+2x-4√x-4√x^3
∫[1,4] x(√x-1)^4dx=∫[1,4](x^3+x+6x^2-4√x^3-4√x^5)dx
=x^4/4+x^2/2+2x^3-4*(2/5)√x^5-4*(2/7)√x^7 |[1,4]
=(64-1/4)+(8-1/2)+(128-2) -8/5(32-1)-(8/7)(128-1)
=198-3/4-248/5-1016/7
=(198-49-145-3/4-3/5-1/7)=4-(105+84+20)/140=351/140
2
∫[-1,3]|3x+1|dx=∫[-1,-1/3](-1-3x)dx+∫[-1/3,3](3x+1)dx
=-x-(3/2)x^2|[-1,-1/3] +(3/2)x^2+x|[-1/3,3]
=(1/3-1/6)-(1-3/2) + (27/2+3)-(1/6-1/3)
=2/3+1/2+3+27/2=17+2/3
3
∫[-3,3] (x^3-x^2-2x+1)dx= x^4/4-x^3/3-x^2+x |[-3,3]
=(-9-9)+3+3=-12