Q. 50 - Calculus

Question

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

${\int }_0^2\sqrt{1+36x}dx$

Step-by-Step Solution

Answer

The function is $f\left(x\right)=4x^{3/2}$ and interval is $\left[a,b\right]=\left[0,2\right]$.

Step 1. Given information.

Consider the given function is ${\int }_0^2\sqrt{1+36x}dx$. 

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 $\begin{array}{rcl}& & {\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx\end{array}$.

Step 3. Find derivative of given function.

Compare given definite integral function with the arc length formula.

$\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_0^2\sqrt{1+36x}dx\\ & =& {\int }_0^2\sqrt{1+{\left(6\sqrt{x}\right)}^2}dx\end{array}$

Therefore the derivative function is $f^{\text{'}}\left(x\right)=\begin{array}{rcl}& & 6\end{array}\begin{array}{rcl}& & \sqrt{x}\end{array}$ and the interval is $\left[0,2\right]$.

Step 4. Find function f x .

The function whose differentiation is $6\sqrt{x}$ is $f\left(x\right)=4x^{3/2}$.

Thus the required function is $f\left(x\right)=4x^{3/2}$ and the interval is $\left[a,b\right]=\left[0,2\right]$.