Find the exact length of the curve, y = x³/3 + 1/4x , 1 ≤ x ≤ 3.

Solution:

Length of the curve, y = f(x) from x =a to x = b is given by:

\( \int_{a}^{b}\sqrt{1+f'(x)^{2}} .dx\)

Given, y = x³/3 + 1/4x , 1 ≤ x ≤ 3

dy/dx = [(1/3)×(3x²)]+[(1/4)×(-1)×x^-2]

dy/dx = f'(x) = x²-(1/4x²)

Length of the curve

=\( \int_{a}^{b}\sqrt{1+f'(x)^{2}} .dx \)

=\( \int_{1}^{3}\sqrt{1+\left ( x^{2}-\frac{1}{4x^{2}} \right )^{2}} .dx \)

=\( \int_{1}^{3}\sqrt{1+\left ( x^{4}-2\times x^{2}\times \frac{1}{4x^{2}}+\frac{1}{16x^{4}} \right )} .dx \)

=\( \int_{1}^{3}\sqrt{1+\left ( x^{4}-\frac{1}{2}+\frac{1}{16x^{4}} \right )} .dx \)

=\( \int_{1}^{3}\sqrt{x^{4}+\frac{1}{2}+\frac{1}{16x^{4}}} .dx \)

=\( \int_{1}^{3}\sqrt{\left ( x^{2}+\frac{1}{4x^{2}}\right)^{2}} .dx \)

=\( \int_{1}^{3}\left ( x^{2}+\frac{1}{4x^{2}}\right).dx \)

=\( \frac{x^{3}}{3}\Biggr|_{1}^{3} +\frac{1}{4} {\frac{x^{-1}}{-1}}\Biggr|_{1}^{3}\)

=\( \frac{1}{3}\left ( 3^{3}-1 \right )-\frac{1}{4}\left (\frac{1}{3}-1 \right )\)

=26/3 + 1/6

=53/6

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Find the exact length of the curve, y = x³/3 + 1/4x , 1 ≤ x ≤ 3.

Summary:

The exact length of the curve, y = x³/3 + 1/4x , 1 ≤ x ≤ 3 is 53/6.