Step by step solution to "Find the exact length of the curve. y=x33+14x,1≤x≤2"

boitshupoO

Answered question

2021-09-17

Find the exact length of the curve.

y = \frac{x_{3}}{3} + \frac{1}{4} x, 1 \leq x \leq 2

tabuordg

Skilled2021-09-18Added 99 answers

The function

f (x) = \frac{x^{3}}{3} + \frac{1}{4 x}

is continuous, differentiable on [1,2] and its derivative

f^{'} (x) = x^{2} - \frac{1}{4 x^{2}}

is continuous on [1,2]
so the length L of the curve is given by the formula:

L = \int_{1}^{2} \sqrt{1 + {[x^{2} - \frac{1}{4 x^{2}}]}^{2}} d x

therefore,

L = \int_{1}^{2} \sqrt{1 + [x^{4} - \frac{1}{2} + \frac{1}{16 x^{4}}]} d x

= \int_{1}^{2} \sqrt{[x^{4} + \frac{1}{2} + \frac{1}{16 x^{4}}]} d x

\int_{1}^{2} \sqrt{{[x^{2} + \frac{1}{4 x^{2}}]}^{2}} d x

\int_{1}^{2} x^{2} + \frac{1}{4 x^{2}} d x

= {[\frac{1}{3} x^{3} - \frac{1}{4 x}]}_{1}^{2}

(\frac{1}{3} 2^{3} - \frac{1}{4 \cdot 2}) - (\frac{1}{3} 1^{3} - \frac{1}{4 \cdot 1})

= \frac{8}{3} - \frac{1}{8} - \frac{1}{3} + \frac{1}{4}

= \frac{64}{24} - \frac{3}{24} - \frac{8}{24} + \frac{6}{24}

L = \frac{59}{24}

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