0 0 votes Probability engineering-mathematics probability + – VS 1.8k views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply Anu007 commented Dec 22, 2017 reply Follow flag m getting 0 as answer 0 0 replyShare sid1221 commented Dec 22, 2017 reply Follow flag 0.03125 0 0 replyShare Anu007 commented Dec 22, 2017 reply Follow flag explain. I'm getting this : $\frac{1}{4}\left ( \frac{1}{16} - \frac{1}{16}\right ) = 0$ 2 2 replyShare sid1221 commented Dec 22, 2017 reply Follow flag o sorry o is correct 0 0 replyShare Shubhanshu commented Dec 22, 2017 reply Follow flag I am also getting 0. 0 0 replyShare srestha commented Dec 22, 2017 reply Follow flag why not 0.03125 0 0 replyShare Please log in or register to add a comment.
Best answer 2 2 votes $\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)dx$ $\int_{-\frac{1}{2}}^{\frac{1}{2}} x^{3}dx$ $\lim_{x->\frac{1}{2}}\left [ \frac{x^{4}}{4} \right ] - \lim_{x->-\frac{1}{2}}\left [ \frac{x^{4}}{4} \right ]$ = $\frac{1}{4}\left ( \frac{1}{16} - \frac{1}{16} \right )=0$ Anu007 answered Dec 22, 2017 • selected Dec 22, 2017 by srestha Anu007 comment Share Follow See all 2 Comments 2 2 Comments reply srestha commented Dec 22, 2017 reply Follow flag $\int_{-\frac{1}{2}}^{\frac{1}{2}} x^{3}dx$ =2 $\int_{\frac{0}{}}^{\frac{1}{2}} x^{3}dx$ =0.03125 is not it? 0 0 replyShare Shubhanshu commented Dec 22, 2017 reply Follow flag This property is applicable when f(x) is even but here it is odd. 3 3 replyShare Please log in or register to add a comment.