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Calculate the arc length of the curve y=x^{2} from x=0 to x=4.

We calculate using the general formula for arc length:

L=\int _{0}^{4}{\sqrt  {1+\left({\frac  {dy}{dx}}\right)^{2}}}\,dx=\int _{0}^{4}{\sqrt  {1+(2x)^{2}}}\,dx. We need to use a trig substitution on this ; namely, let 2x=\tan \theta , so 2\,dx=\sec ^{2}\theta \,d\theta . Calculate the antiderivative first:

\int {\sqrt  {1+4x^{2}}}\,dx=\int {\sqrt  {1+\tan ^{2}\theta }}\cdot {\frac  {1}{2}}\sec ^{2}\theta \,d\theta =\int {\frac  {1}{2}}\sec ^{3}\theta \,d\theta ={\frac  {1}{4}}\left(\tan \theta \,\sec \theta +\ln |\tan \theta +\sec \theta |\right). We substitute for x to get \int {\sqrt  {1+4x^{2}}}\,dx={\frac  {1}{4}}\left(2x{\sqrt  {1+4x^{2}}}+\ln |2x+{\sqrt  {1+4x^{2}}}|\right)+C.

Then L=\int _{0}^{4}{\sqrt  {1+4x^{2}}}\,dx=\left.{\frac  {1}{4}}\left(2x{\sqrt  {1+4x^{2}}}+\ln |2x+{\sqrt  {1+4x^{2}}}|\right)\right|_{{x=0}}^{4}={\frac  {1}{4}}\left(8{\sqrt  {65}}+\ln(8+{\sqrt  {65}})\right).