## anonymous one year ago FInd the length of curve y=x^3+(1/(12x)) on interval [1/2,2] I get that I take the derivative and square it. and plug it in to sqrt(1+y'^2) but what confuses me is do I do the inverse of derivative to equal out. also can I just take the square root of everything inside and then take the integral?

1. freckles

you can take the square root of the thing inside if the inside itself can be written as a square but depending on the limits you might have to adjust your sign of the integrand

2. freckles

so for this one what do you get as y'?

3. anonymous

I get $\sqrt{1+((3x^2-(1/12x^2))^2}$ so could I basically just cancel out the square root

4. freckles

no no

5. freckles

the whole thing inside has to be written as a square for you to do that

6. freckles

like so: $1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^2}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^2} \\ (3x^2+\frac{1}{12x^2})^2$

7. freckles

$\sqrt{(3x^2+\frac{1}{12x^2})^2}=3x^2+\frac{1}{12x^2} \text{ for all } x \neq 0$

8. freckles

which we don't have to worry about 0 since we are looking at [1/2,2]

9. freckles

so this is what you have: $\int\limits_{1/2}^{2} (3x^2+\frac{1}{12x^2}) dx$

10. freckles

11. freckles

$1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x\color{red}{^4}}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x\color{red}{^4}} \\ (3x^2+\frac{1}{12x^2})^2$

12. freckles

by the way the whole sign thing I mentioned earlier...pretend we have: $\int\limits_{-1}^2 \sqrt{x^2} dx \\ \text{ well } \sqrt{x^2}=x \text{ if } x \ge 0 \\ \text{ and } \sqrt{x^2}=-x \text{ if } x<0 \\ \text{ so } \\ \int\limits_{-1}^2 \sqrt{x^2} dx=\int\limits_{-1}^0 (-x )dx+\int\limits_0^2 x dx$

13. anonymous

So in reality there is no square root for this problem? because the two integrals are equal to it?

14. freckles

yeah we got rid of the square root by writing previous integrand as $3x^2+\frac{1}{12x^2}$

15. anonymous

didn't you forget the plus 1?

16. freckles

remember this: $1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^4}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^4} \\ (3x^2+\frac{1}{12x^2})^2$

17. freckles

those 4 expressions are equal expressions

18. freckles

$\sqrt{1+(3x^2-\frac{1}{12x^2})^2} =\\ \sqrt{(3x^2+\frac{1}{12x^2})^2}$

19. freckles

hey @Empty and @IrishBoy123 I have to leave can you help @Greenwalrus further

20. anonymous

ah yeah I see!

21. anonymous

help :(

22. anonymous

since I don't have a negative I can just ignore having two integrals and have one right