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Determine which value best approximates the length of the arc represented by the integral 0π/41+(sec2x)2dx\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) π\pi, (d) 4, (e) 3.


Answered 2 years ago
Answered 2 years ago
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s=0π41+[sec2x]2 dxab1+[f(x)]2 dx=0π41+[sec2x]2 dx\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=0a=0, b=π4b=\frac{\pi}{4} and f(x)=sec2xf'(x)=\sec^2 x. Let's find f(x)f(x) and make a sketch of f(x)f(x).

f(x)=sec2xf(x) dx=sec2x dxf(x)=tanx+C\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=0C=0. Therefore, f(x)=tanxf(x)=\tan x. From the sketch, we can estimate that the length is equal to 1.

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