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`int_(0)^(4)1/(1+sqrtx)dx`=…..A. `log((e^(4))/6)`B. `log((e^(4))/9)`C. `log((e^(4))/9)`D. `log((e^(4))/4)` |
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Answer» Correct Answer - C Let `I=int_(0)^(4)1/(1+sqrtx)dx` Putting, `1+sqrtx=t!1/(2sqrtx)dx=dt` `!dx=2sqrtxdt!dx=2(t-1)dt` at `x=0,t=1` and x=4,t=3 Now, `I=int_(1)^(3)(2(t-1)dt)/t=2int_(1)^(3)(1-1/t)dt` `=2[t-logabst]_(1)^(3)` `=2[(3-log3)-(1-log1)]` `=2[2-log3](becauselog1=0)` `=4+2log(1/3)=4loge+log(1/9)` `=loge^(4)+log(1/9)=log(e^(4)/9)` |
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