Question

calculate the arc length over the given interval.

1. \\( y = 6 - 2x \\), \\( 1, 3 \\)

© 2018 flamingo math™

Explanation:

Step1: Recall Arc Length Formula

The arc length \( L \) of a function \( y = f(x) \) from \( x = a \) to \( x = b \) is given by \( L=\int_{a}^{b}\sqrt{1+(f'(x))^{2}}dx \). First, find the derivative of \( y = 6 - 2x \). The derivative \( y'=f'(x)=\frac{d}{dx}(6 - 2x)=- 2 \).

Step2: Substitute into Arc Length Formula

Substitute \( f'(x)=-2 \) into the formula: \( L=\int_{1}^{3}\sqrt{1+(-2)^{2}}dx=\int_{1}^{3}\sqrt{1 + 4}dx=\int_{1}^{3}\sqrt{5}dx \).

Step3: Evaluate the Integral

Since \( \sqrt{5} \) is a constant, the integral of a constant \( C \) from \( a \) to \( b \) is \( C(b - a) \). So, \( L=\sqrt{5}(3 - 1)=2\sqrt{5} \). Alternatively, notice that \( y = 6 - 2x \) is a straight line, so the arc length is just the distance between the two points. When \( x = 1 \), \( y=6-2(1)=4 \); when \( x = 3 \), \( y=6-2(3)=0 \). The distance between \( (1,4) \) and \( (3,0) \) is \( \sqrt{(3 - 1)^{2}+(0 - 4)^{2}}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5} \), which matches the integral result.

Answer:

\( 2\sqrt{5} \)