Can someone check thru my working? I cant find mymistake... The ans is b =4
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you integrated wrong. $f(x)=\begin{cases} \sqrt{\dfrac{x-1}{12}}&1\leq x\leq b\\ 0&\mbox{else} \end{cases}$ $\displaystyle{\int_1^b}f(x)~dx=1$ $\displaystyle{\int_1^b}\sqrt{\dfrac{x-1}{12}}~dx=1$ $\dfrac{(b-1)^{3/2}}{3\sqrt{3}}=1$ $b=4$
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