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Evaluate : (i) int(-pi//2)^(pi//2)|sin...

Evaluate :
`(i) int_(-pi//2)^(pi//2)|sinx|dx` `(ii) int_(-1)^(1)e^(|x|)dx` `(iii) int_(-2)^(1)|2x+1|dx`.

Text Solution

Verified by Experts

`(i)` Clearly, `|sinx|` is an even function of `x`.
`:.int_(-pi//2)^(pi//2)|sinx|dx=2int_(0)^(pi//2)|sinx|dx`
`=2int_(0)^(pi//2)sinxdx[:'sinx ge 0, "when" 0 le x le (pi)/(2)]`
`=[-2cosx]_(0)^(pi//2)=2`.
`(ii)` Clearly, `e^(|x|)` is an even function of `x`.
`:.int_(-1)^(1)e^(|x|)dx=2int_(0)^(1)e^(|x|)dx`
`=2int_(0)^(1)e^(x)dx[:'|x|=x,"when" 0 le x le 1]`
`=[2e^(x)]_(0)^(1)=(2e-2)=2(e-1)`.
`(iii) [-2 le x lt -(1)/(2)implies2x+1 lt 0]` and `[-(1)/(2) le x le 1implies2x+1 ge 0]`
`:.int_(-2)^(1)|2x+1|dx=int_(-2)^(-1//2)|2x+1|dx+int_(-1//2)^(1)|2x+1|dx`
`=int_(-2)^(-1//2)-(2x+1)dx+int_(-1//2)^(1)(2x+1)dx`
`=[-x^(2)-x]_(-2)^(-1//2)+[x^(2)+x]_(-1//2)^(1)`
`=(-(1)/(4)+(1)/(2))-(-4+2)+[2-((1)/(4)-(1)/(2))]`
`=(1)/(4)+2+(9)/(4)=(9)/(2)`.
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