Question

求定积分，只有两小题，必采纳!

Answer 1

解：
∫（0,π）√cos²xdx

=∫（0,π）|cosx|dx
=∫（0,π/2）cosxdx-∫（π/2,π）cosxdx
=1+1
=2
∫（-2,3）|x²-2x-3|dx
=∫（-2,3）|(x-3)(x+1)|dx
=∫（-2,-1）|(x-3)(x+1)|dx+∫（-1,3）|(x-3)(x+1)|dx
=∫（-2,-1）(x-3)(x+1)dx-∫（-1,3）(x-3)(x+1)dx
=[（1/3）x^3-x^2-3x]|(-2,-1)+[（-1/3）x^3+x^2+3x]|(-1,3)
=7/3+32/3
=13

Answer 2

∫<0,π>√cos²xdx

=∫<0,π>|cosx|dx
=∫<0,π/2>cosxdx-∫<π/2,π>cosxdx
=1+1
=2
∫<-2,3>|x²-2x-3|dx
=∫<-2,3>|(x-3)(x+1)|dx
=∫<-2,-1>|(x-3)(x+1)|dx+∫<-1,3>|(x-3)(x+1)|dx
=∫<-2,-1>(x-3)(x+1)dx-∫<-1,3>(x-3)(x+1)dx
=-5

Answer 3

2