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Match the following lists: |
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Answer» `=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)` |
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