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If`Isum_(k=1)^(98)int_k^(k+1)(k+1)/(x(x+1))dx ,t h e n:``I(log)_e 99``I >(log)_e 99`(d) `I |
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Answer» Correct Answer - B::D `I=sum_(k=1)^(98)int_(k)^(k+1)((k+1))/(x(x+1))dx` Clearly , `Igtsum_(k=1)^(98)int_(k)^(k+1)((k+1))/((x+1)^(2))dx` `rArrIgtsum_(k=1)^(98)(k+1)int_(k)^(k+1)(1)/((x+1))dx` `rArrIgtsum_(k=1)^(98)(-(k+1))[(1)/(k+2)-(1)/(k+21)]rArrIgtsum_(k=1)^(98)(1)/(k+2)` `rArrIgt (1)/(3)+ ...+(1)/(100)gt(98)/(100)rArrIgt(49)/(50)` Also , `Igtsum_(k=1)^(98)int_(k)^(k+1)(k+1)/(x(k+1))dx=sum_(k=1)^(98)[log_(e)(k+1)-log_(e)k]` `Igtlog_(e)99` |
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