Question

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

${\int }_0^{\mathrm{\pi }}\sqrt{1+{\sin }^{2}x}dx$

Short answer

The function is$f\left(x\right)=-\cos x$and$\left[a,b\right]=\left[0,\mathrm{\pi }\right]$.

Step-by-step solution

Step 1. Given information.

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

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.

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

Therefore the function$f^{\text{'}}\left(x\right)=\sin x$androle="math" localid="1649235589478" $\left[a,b\right]=\left[0,\mathrm{\pi }\right]$.

Step 4. Find function fx.

The function whose differentiation is $\sin x$is role="math" localid="1649235561134" $f\left(x\right)=-\cos x$.

Thus the required function is $f\left(x\right)=-\cos x$and$\left[a,b\right]=\left[0,\mathrm{\pi }\right]$.