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Find \(\int_1^2 \frac{12 \,logx}{x} \,dx\).(a) -12 log2(b) 24 log2(c) 12 log2(d) 24 log4This question was posed to me in an interview.This key question is from Evaluation of Definite Integrals by Substitution in portion Integrals of Mathematics – Class 12 |
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Answer» RIGHT choice is (b) 24 log2 To explain I WOULD SAY: I=\(\int_1^2 \frac{12 logx}{x} \,dx\) Let logx=t Differentiating w.r.t x, we get \(\frac{1}{x} \,dx=dt\) The new limits When x=1,t=0 When x=2,t=log2 \(\int_1^2 \frac{12 logx}{x} dx=12\int_0^{log2} \,t \,dt\) =\(12[t^2]_0^{log2}=12((log2)^2-0)\) =12 log4=24 log2(∵(log2)^2=log2.log2=log4=2 log2) |
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