Question

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

${\int }_{-2}^5\sqrt{1+9e^{6x}}dx$

Short answer

The function is$f\left(x\right)=e^{3x}$and$\left[a,b\right]=\left[-2,5\right]$.

Step-by-step solution

Step 1. Given information.

Consider the given function is${\int }_{-2}^5\sqrt{1+9e^{6x}}dx$.

Step 2. Use arc length formula. 

The formula for a function to find the arc length from $x=a$to $x=b$is given byrole="math" localid="1649233700729" ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find derivative of given function.

Compare given definite integral function with the arc length formula.

$\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_{-2}^5\sqrt{1+9e^{6x}}dx\\ & =& {\int }_{-2}^5\sqrt{1+{\left(3e^{3x}\right)}^2}dx\end{array}$

Therefore the function $f^{\text{'}}\left(x\right)$will be$f^{\text{'}}\left(x\right)=3e^{3x}$and$\left[a,b\right]=\left[-2,5\right]$.

Step 4. Find function f(x).

The function whose differentiation is $3e^{3x}$is role="math" localid="1649234040169" $f\left(x\right)=e^{3x}$.

Thus the required function is$f\left(x\right)=e^{3x}$and$\left[a,b\right]=\left[-2,5\right]$.