# Find the exact length of the curve. y=x_3/3+1/4x,1\leq x\leq 2

Find the exact length of the curve.
$$\displaystyle{y}=\frac{{x}_{{3}}}{{3}}+\frac{{1}}{{4}}{x},{1}\leq{x}\leq{2}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

tabuordg
The function $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{3}}}}{{{3}}}}+{\frac{{{1}}}{{{4}{x}}}}$$ is continuous, differentiable on [1,2] and its derivative
$$\displaystyle{f}'{\left({x}\right)}={x}^{{2}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}$$ is continuous on [1,2]
so the length L of the curve is given by the formula:
$$\displaystyle{L}={\int_{{1}}^{{2}}}\sqrt{{{1}+{\left[{x}^{{2}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}\right]}^{{2}}}}{\left.{d}{x}\right.}$$
therefore,
$$\displaystyle{L}={\int_{{1}}^{{2}}}\sqrt{{{1}+{\left[{x}^{{4}}-{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{16}{x}^{{4}}}}}\right]}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{1}}^{{2}}}\sqrt{{{\left[{x}^{{4}}+{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{16}{x}^{{4}}}}}\right]}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{1}}^{{2}}}\sqrt{{{\left[{x}^{{2}}+{\frac{{{1}}}{{{4}{x}^{{2}}}}}\right]}^{{2}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{1}}^{{2}}}{x}^{{2}}+{\frac{{{1}}}{{{4}{x}^{{2}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[{\frac{{{1}}}{{{3}}}}{x}^{{3}}-{\frac{{{1}}}{{{4}{x}}}}\right]}_{{1}}^{{2}}}$$
$$\displaystyle{\left({\frac{{{1}}}{{{3}}}}{2}^{{3}}-{\frac{{{1}}}{{{4}\cdot{2}}}}\right)}-{\left({\frac{{{1}}}{{{3}}}}{1}^{{3}}-{\frac{{{1}}}{{{4}\cdot{1}}}}\right)}$$
$$\displaystyle={\frac{{{8}}}{{{3}}}}-{\frac{{{1}}}{{{8}}}}-{\frac{{{1}}}{{{3}}}}+{\frac{{{1}}}{{{4}}}}$$
$$\displaystyle={\frac{{{64}}}{{{24}}}}-{\frac{{{3}}}{{{24}}}}-{\frac{{{8}}}{{{24}}}}+{\frac{{{6}}}{{{24}}}}$$
$$\displaystyle{L}={\frac{{{59}}}{{{24}}}}$$