\[y=f(x)\]
from \[(a,f(a))\]
to \[(b, f(b))\]
is \[L= \int^b_a \sqrt{1+ (\frac{df}{dx})^2}dx\]
.The length of the curve
\[y=x^2\]
between \[x=0, \: 1\]
is
\[L= \int^1_0 \sqrt{1+ (\frac{d(x^2}{dx})^2}dx = \int^1_0 \sqrt{1+4x^2} dx \]
.Substitute
\[2x=tan u \2 \frac{dx}{du}= sec^2 u \rightarrow dx=\frac{1}{2}sec^2udu\]
.When
\[x=1, \: 2=tanu \rightarrow u=tan^{-1}(\frac{1}{2}) \]
and when \[x=0, \: 0=tanu \rightarrow u=0\]
.\[\begin{equation} \begin{aligned} L &= \int^{tan^{-1}(\frac{1}{2})}_0 \sqrt{1+tan^2u} \frac{1}{2}sec^2udu \\ &= \int^{tan^{-1}(\frac{1}{2})}_0 \sqrt{sec^2u} \frac{1}{2}sec^2udu \\ &= \frac{1}{2} \int^{tan^{-1}(\frac{1}{2})}_0 secu (1+tan^2u)du \\ &= \frac{1}{2} \int^{tan^{-1}(\frac{1}{2})}_0 secu+secutan^2udu \\ &= \frac{1}{2} [ln(secu+tanu)+\frac{1}{2}tan^2u]^{tan^{-1}(\frac{1}{2})}_0 \\ &= \frac{1}{2} [ln(\sqrt{1+tan^2u}+tanu)+\frac{1}{2}tan^2u]^{tan^{-1}(\frac{1}{2})}_0 \\ &= \frac{1}{2} ((ln(\sqrt{5})+4)-(ln(1)+0))= \frac{1}{2} (ln(\sqrt{5}+4) \end{aligned} \end{equation}\]