Q. 49 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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 f x .

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 $\left[a,b\right]=\left[2,3\right]$.