Question
Show that the arc length of $y=\ln (f(x))$ for $a \leq x \leq b$ is$\int_{a}^{b} \frac{\sqrt{f(x)^{2}+f^{\prime}(x)^{2}}}{f(x)} d x$
Step 1
Step 1: We start with the formula for arc length, which is given by \[L=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x\] Show more…
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Show that if $0 \leq f^{\prime}(x) \leq 1$ for all $x,$ then the arc length of $y=f(x)$ over $[a, b]$ is at most $\sqrt{2}(b-a) .$ Show that for $f(x)=x$ the arc length equals $\sqrt{2}(b-a)$
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For the function f(x) = 1/4e^x + e^-x, prove that the arc length on any interval has the same value as the area under the curve. f(x) = 1/4e^x + e^-x => f'(x) = => 1 + [f'(x)]^2 = 1 + = ( )^2 = [f(x)]^2 The arc length of the curve y = f(x) on the interval [a, b] is L = ∫_a^b sqrt(1 + [f'(x)]^2) dx = ∫_a^b sqrt([f(x)]^2) dx = ∫_a^b f(x) dx, which is the area under the curve y = f(x) on the interval [a, b].
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