I can't seem to find the area bounded by the s-curves. Help?
Find the area between $y=\frac{-1}{6\sin x}+3$ and $y=\frac{1}{6\sin x}+3$
0
$\begingroup$
calculus
integration
derivatives
area
-
0yup i meant that – 2017-02-27
-
0Is that a problem in the book or you just made it by yourself? – 2017-02-27
-
0I graphed it on an application but I can't get its area – 2017-02-27
-
1Of course, you can't find the area since the graphs don't have intersection?. – 2017-02-27
-
0The intersection is x=(pi)(n)? – 2017-02-27
-
1No. The value of $\sin x$ is uniquely determined. – 2017-02-27
1 Answers
1
You can't find the area bounded by the curves since they do not intersect for all $x\in \mathbb{R}$. See the graphs I've plotted below on Desmos Graphing Calculator:
-
0I'm a user of desmos also. But I cant do it by pasting the commands. How did you do that. – 2017-02-27
-
0Are you using the Print Screen Command? – 2017-02-27
-
0@ΘΣΦGenSan Yes, I used "Print Screen Command". Good question. – 2017-02-27
-
0Thank you. I will try it now. – 2017-02-27
-
1Success, thanks for your help. I know how do it now. (+1) – 2017-02-27
