# What is the arclength of #f(x)=2-3x # in the interval #[-2,1]#?

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To find the arc length of a function ( f(x) ) in the interval ([a, b]), you use the formula:

[ L = \int_a^b \sqrt{1 + (f'(x))^2} , dx ]

Given ( f(x) = 2 - 3x ) and the interval ([-2, 1]), first find ( f'(x) ), which is -3. Then, plug this into the arc length formula and integrate from -2 to 1:

[ L = \int_{-2}^1 \sqrt{1 + (-3)^2} , dx ]

Simplify the expression:

[ L = \int_{-2}^1 \sqrt{10} , dx ]

Integrate (\sqrt{10}) from -2 to 1:

[ L = \sqrt{10} \int_{-2}^1 dx ]

Evaluate the integral:

[ L = \sqrt{10} \left[ x \right]_{-2}^1 ]

Substitute the limits:

[ L = \sqrt{10} \left(1 - (-2)\right) ]

[ L = \sqrt{10} \times 3 ]

[ L = 3\sqrt{10} ]

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