Question

Find the length of the curve for the interval:

Answer 1

\[y=\frac{x^4}{4}+\frac{1}{8x^2}\]
interval is [1,3]

the length of the curve can be found by using the following formula
\[L=\int\limits_{1}^{3}\sqrt{1+({y')}^2} dx\]

Answer 2

so what is y'?

Answer 3

Do you know how to find y'?

Answer 4

would it be by adding x^4/4+1/8x^2 together by finding the common denom, and the taking the derivative?

Answer 5

\[y'=4 \cdot \frac{x^3}{4}+\frac{1}{8} (x^{-2})'=x^3+\frac{1}{8}(-2x^{-2-1})=x^3-\frac{2}{8x^3}=x^3-\frac{1}{4x^3}\]

Answer 6

\[(y')^2=(x^3-\frac{1}{4x^3})^2=x^6-2x^3 \frac{1}{4x^3}+\frac{1}{16x^6}\]
\[(y')^2=x^6-\frac{1}{2}+\frac{1}{16x^6}\]
now plug this into the formula

Answer 7

so we have
\[\int\limits_{1}^{3}\sqrt{1+x^6-\frac{1}{2}+\frac{1}{16x^6}} dx=\int\limits_{1}^{3}\sqrt{x^6+\frac{1}{2}+\frac{1}{16x^6}} dx\]

Answer 8

\[\int\limits_{1}^{3}\sqrt{\frac{16x^{12}+8x^6+1}{16x^6} }dx\]

Answer 9

\[\int\limits_{1}^{3}\frac{\sqrt{16x^{12}+8x^6+1}}{4x^3} dx\]

Answer 10

now that thing that is left under the radical is that a perfect square

Answer 11

\[(4x^6)^2+2 \cdot 4x^6 \cdot 1 +(1)^2=(4x^6+1)^2\]

Answer 12

\[\int\limits_{1}^{3}\frac{4x^6+1}{4x^3} dx=\int\limits_{1}^{3}(x^3+\frac{1}{4}x^{-3}) dx\]

Answer 13

You can finish this right?

Answer 14

Yeah, I do believe so

Answer 15

hopefully i didn't make a mistake
it is past my bedtime

Answer 16

you did it right :)...very nice

Answer 17

would i use u sub

Answer 18

no don't use u-sub

Answer 19

thats why i was having a problem, lol

Answer 20

\[[\frac{x^4}{4}+\frac{1}{4} \cdot \frac{x^{-3+1}}{-3+1}]_1^3\]
now you got it?

Answer 21

but what method did you use?

Answer 22

I just used \[\int\limits_{}^{}x^n dx=\frac{x^{n+1}}{n+1}+C , n \neq -1\]

Answer 23

Thanks for the help!