Concept:

The length (L) of curve y(x) in interval (a,b) is given as:

\(L = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]dx } \)   -----(1)

Calculation:

Given:

y(x) = x^3/2

(a, b) = (0, 1)

\(\frac{{dy\left( x \right)}}{{dx}} = \frac{3}{2}\;{x^{\frac{1}{2}}}\)

From equation (1)

\(I = \mathop \smallint \limits_a^b \sqrt {\left[ {1 + \;\left( {\frac{9}{4}x} \right)} \right]} dx\)

\( = \frac{1}{2}\mathop \smallint \limits_0^1 \sqrt {\left( {4 + 9x} \right).dx} \)

Let, u = 4 + 9x

\(\frac{{\partial u}}{{\partial x}} = 9 = \frac{1}{9}\smallint du = dx\)

\(L = \left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\mathop \smallint \limits_4^{13} \left( {\sqrt u } \right).du\)

\(L = \;\left( {\frac{1}{2}} \right)\left( {\frac{1}{9}} \right)\left( {\frac{2}{3}} \right){\left[ {{4^{\frac{3}{2}}}} \right]_4}^{13}\)

\(L = \frac{1}{{27}}[\left( {13{)^{\frac{3}{2}}} - {{\left( 4 \right)}^{\frac{3}{2}}}} \right]\)

\(L = \frac{1}{{27}}\left( {{{\left( {13} \right)}^{\frac{3}{2}}} - 8} \right)\)