## Related questions with answers

Determine which value best approximates the length of the arc represented by the integral $\int_{0}^{\pi / 4} \sqrt{1+\left(\sec ^{2} x\right)^{2}} d x$. (Make your selection on the basis of a sketch of the arc and not by performing any calculations.) (a) -2, (b) 1, (c) $\pi$, (d) 4, (e) 3.

Solution

Verified$\begin{align*} s&=\int_0^{\frac{\pi}{4}} \sqrt{1+\left[ \sec^2 x \right]^2}\ dx \\ \int_a^b \sqrt{1+\left[ f'(x) \right]^2}\ dx&=\int_0^{\frac{\pi}{4}} \sqrt{1+\left[ \sec^2 x \right]^2}\ dx \\ \end{align*}$

Hence, we see that $a=0$, $b=\frac{\pi}{4}$ and $f'(x)=\sec^2 x$. Let's find $f(x)$ and make a sketch of $f(x)$.

$\begin{align*} f'(x)&=\sec^2 x \int f'(x)\ dx=\int \sec^2 x\ dx \\ f(x)&=\tan x+C \end{align*}$

Suppose, without loss of generality, that it is $C=0$. Therefore, $f(x)=\tan x$. From the sketch, we can estimate that the length is equal to 1.

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