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Let `f(x)=int_(0)^(x)(e^(t))/(t)dt(xgt0),` then `e^(-a)[f(x+1)-f(1+a)]=`A. `int_(0)^(x)=(e^(t))/((t+a))dt`B. `int_(1)^(x)(e^(t))/(t+a)dt`C. `e^(-a)int_(1+a)^(x+a)(e^(t))/(t)dt`D. `int_(0)^(x)(e^(t-a))/((t+a))dt` |
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Answer» Correct Answer - B::C `e^(-a)[f(x+a)-f(1+a)]` `" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)-int_(0)^(1+a)(e^(t))/(t)dt]` `" "=e^(-a)[int_(0)^(x+a)(e^(t).dt)/(t)+int_(1+a)^(0)(e^(t))/(t)dt]` `" "=e^(-a)[int_(1+a)^(x+a)(d^t.dt)/(t)]` `" "=e^(-a)int_(1)^(x)(e^(y+a))/(y+a).dy" "("Put, t"=y+a, dt=dy)` `" "=int_(1)^(x)(e^(y).dy)/(t+a)` `" "=int_(1)^(x)(e^(t).dt)/(t+a)` |
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