Evaluate `int_(0)^(1.5) x[x^2] dx`, where [.] denotes the greatest integer function

Answer

`ator;btop;ctos;dtoq`

Explanation

`lim_(nto oo) [(int_(0)^(2)(1+t/(n+1))^(n)dt)/(n+1)]=lim_(n to oo) [(1+t/(n+1))^(n-1)]_(0)^(2)`
`=lim_(nto oo) (1+2/(n+1))^(n+1)-1`
`=e^(2)-1`
`f'(x)=f(x)impliesf(x)=Ce^(x)` and since `f(0)=1, 1=f(0)=C`
Therefore `f(x)=e^(x)` and hence `g(x)=x^(2)-e^(x)`.
Thus, `int_(0)^(1)f(x)g(x)dx=int_(0)^(1)(x^(2)e^(x)-e^(2x))dx`
`=x^(2)e^(x)|_(0)^(1)-2int_(0)^(1)xe^(x) dx=int_(0)^(1)(x^(2)e^(x)-e^(2x))dx`
`=x^(2)e^(x)|._(0)^(1)-2int_(0)^(1)xe^(x)dx-(e^(2x))/2|_(0)^(1)`
`=(e-0)-2xe^(x)|._(0)^(1)+2e^(x)|_(0)^(1)+2e^(x)|_(0)^(1)-1/2(e^(2)-1)`
`=(e-0)-2e+2e-2-1/2(e^(2)-1)`
`=e-1/2e^(2)-3/2`
`I=int_(0)^(1)e^(e^(x))(1+xe^(x))dx`
Let `e^(x)=t`
`:. int_(1)^(e)(1+tlogt)(dt)/t int_(1)^(e)e^(t)(1/t +logt)dt`
`=[e^(t)logt]_(1)^(e)=e^(e)`
d. `L=lim_(kto0)(int_(0)^(k)(1+sin2x)^(1/x)dx)/k` (form `0/0`)
`=lim_(kto0) (1+sin2k)^(1/k)`
`=e^(lim_(kto0)1/k(sin2k))=e^(2)`