What is the arc length of #f(x)=2-3x# on #x in [-2,1]#?

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Answer 1 · 2016-01-26T10:53:51

#3sqrt10#

#f'(x) = -3#

#dL = sqrt{1 + (f'(x))^2} dx#

#= sqrt{1 + (-3)^2} dx#

#= sqrt{10} dx#

#L = int_{-2}^1 sqrt{10} dx#

#= 3sqrt{10}#

Alternatively, since this is a straight line, we can use Pythagoras' Theorem.

The horizontal is #1-(-2)=3# units.
The vertical is #f(1)-f(-2)=(-1)-11=-12# units.

Using Pythagoras' Theorem, we have the length of the diagonal as

#sqrt{3^2+12^2} = 3sqrt10#