How do you set up an integral from the length of the curve $y = \frac{1}{x}, 1 \leq x \leq 5$ $y=1/x, 1<=x<=5$ ?

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Answer 1 · 2017-02-25T18:02:33

$\int_{1}^{5} \sqrt{1 + \frac{1}{x^{4}}}$ $int_1^5sqrt(1+1/x^4)$

The length of the curve of $y$ $y$ on $a \leq x \leq b$ $alt=xlt=b$ is equal to:

$L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}}$ $L=int_a^bsqrt(1+(dy/dx)^2)$

For $y = \frac{1}{x}$ $y=1/x$ , we see that $y = x^{- 1}$ $y=x^-1$ so $\frac{d y}{d x} = - x^{- 2} = - \frac{1}{x^{2}}$ $dy/dx=-x^-2=-1/x^2$ . Then:

$L = \int_{1}^{5} \sqrt{1 + {(- \frac{1}{x^{2}})}^{2}} = \int_{1}^{5} \sqrt{1 + \frac{1}{x^{4}}}$ $L=int_1^5sqrt(1+(-1/x^2)^2)=int_1^5sqrt(1+1/x^4)$

Which can be plugged into a calculator for a result of $4.15145$ $4.15145$ .

How do you set up an integral from the length of the curve y=1/x, 1<=x<=5y=1x,1≤x≤5y=1/x, 1<=x<=5?