Question

Each definite integral represents the arc length of a function f(x) on an interval [a, b]. Determine the function and interval.

${\int }_2^3\sqrt{\frac{x^4+1}{x^4}}dx$

Short answer

The function is$f\left(x\right)=\frac{1}{x}$and$\left[a,b\right]=\left[2,3\right]$.

Step-by-step solution

Step 1. Given information.

Consider the given integral is ${\int }_2^3\sqrt{\frac{x^4+1}{x^4}}dx$.

Step 2. Use arc length formula. 

The formula for a function to find the arc length from $x=a$to $x=b$is given by${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find derivative of given function.

Compare given definite integral function with the arc length formula.

$\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_2^3\sqrt{\frac{x^4+1}{x^4}}dx\\ & =& {\int }_2^3\sqrt{1+{\left(-\frac{1}{x^2}\right)}^2}dx\end{array}$

Therefore the derivative function is$f^{\text{'}}\left(x\right)=\left(-\frac{1}{x^2}\right)$and limit is$\left[a,b\right]=\left[2,3\right]$.

Step 4. Find function fx.

The function whose differentiation is $\left(-\frac{1}{x^2}\right)$is $f\left(x\right)=\frac{1}{x}$.

Thus the required function is$f\left(x\right)=\frac{1}{x}$and$\left[a,b\right]=\left[2,3\right]$.