Q. 52 - Calculus

Question

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

${\int }_0^1\sqrt{1+9x^4}dx$ 

Step-by-Step Solution

Answer

The function is $f\left(x\right)=x^3$ and interval is $\left[a,b\right]=\left[0,1\right]$.

Step 1. Given information.

Consider the given function is ${\int }_0^1\sqrt{1+9x^4}dx$.

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.

 $\begin{array}{rcl}{\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx& =& {\int }_0^1\sqrt{1+9x^4}dx\\ & =& {\int }_0^1\sqrt{1+{\left(3x^2\right)}^2}dx\end{array}$

Therefore the derivative function will be $f^{\text{'}}\left(x\right)=3x^2$ and the interval is $\left[a,b\right]=\left[0,1\right]$.

Step 4. Find function f x .

The function whose differentiation is $3x^2$ is $f\left(x\right)=x^3$.

Thus the required function is $f\left(x\right)=x^3$ and $\left[a,b\right]=\left[0,1\right]$.