## Length of a Curve Example

The length of a curve
$y=f(x)$
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{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}