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1. Find the length of the curve
![{\displaystyle y=x^{\frac {3}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68eb525f9b132b04521a5e182db7c319d9c88e05)
from
![{\displaystyle x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953917eaf52f2e1baad54c8c9e3d6f9bb3710cdc)
to
![{\displaystyle x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee42176e76ae6b56d68c42ced807e08b962a2b54)
.
:
Let
![{\displaystyle u=1+{\frac {9}{4}}x;\quad du={\frac {9}{4}}dx;\quad dx={\frac {4}{9}}du}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32617c2ed72c43f647725e7b2248343d7799d388)
Then
![{\displaystyle {\begin{aligned}s&=\int \limits _{u(0)}^{u(1)}{\frac {4}{9}}{\sqrt {u}}\,du\\&={\frac {4}{9}}\cdot {\frac {2}{3}}{\Big [}u^{\frac {3}{2}}{\Big ]}_{u(0)}^{u(1)}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}x\right)^{\frac {3}{2}}\right]_{0}^{1}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}\right)^{\frac {3}{2}}-1\right]\\&=\mathbf {\frac {13{\sqrt {13}}-8}{27}} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1698056f2bb851198b40eaeca57953f876ed6fce)
:
Let
![{\displaystyle u=1+{\frac {9}{4}}x;\quad du={\frac {9}{4}}dx;\quad dx={\frac {4}{9}}du}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32617c2ed72c43f647725e7b2248343d7799d388)
Then
![{\displaystyle {\begin{aligned}s&=\int \limits _{u(0)}^{u(1)}{\frac {4}{9}}{\sqrt {u}}\,du\\&={\frac {4}{9}}\cdot {\frac {2}{3}}{\Big [}u^{\frac {3}{2}}{\Big ]}_{u(0)}^{u(1)}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}x\right)^{\frac {3}{2}}\right]_{0}^{1}\\&={\frac {8}{27}}\left[\left(1+{\frac {9}{4}}\right)^{\frac {3}{2}}-1\right]\\&=\mathbf {\frac {13{\sqrt {13}}-8}{27}} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1698056f2bb851198b40eaeca57953f876ed6fce)
2. Find the length of the curve
![{\displaystyle y={\frac {e^{x}+e^{-x}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a7c864ed63fa34292d8eb0685bdbdaedfd3dc14)
from
![{\displaystyle x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953917eaf52f2e1baad54c8c9e3d6f9bb3710cdc)
to
![{\displaystyle x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee42176e76ae6b56d68c42ced807e08b962a2b54)
.
:
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {d}{dx}}\left({\frac {e^{x}+e^{-x}}{2}}\right)\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {e^{x}-e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+{\frac {e^{2x}-2+e^{-2x}}{4}}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\frac {e^{2x}+2+e^{-2x}}{4}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\left({\frac {e^{x}+e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\frac {e^{x}+e^{-x}}{2}}dx\\&={\frac {e^{x}-e^{-x}}{2}}{\bigg |}_{0}^{1}\\&=\mathbf {\frac {e-{\frac {1}{e}}}{2}} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c37fa7e5bde769230d12416141f2e0b9080ed7f)
:
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {d}{dx}}\left({\frac {e^{x}+e^{-x}}{2}}\right)\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+\left({\frac {e^{x}-e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\sqrt {1+{\frac {e^{2x}-2+e^{-2x}}{4}}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\frac {e^{2x}+2+e^{-2x}}{4}}}dx\\&=\int \limits _{0}^{1}{\sqrt {\left({\frac {e^{x}+e^{-x}}{2}}\right)^{2}}}dx\\&=\int \limits _{0}^{1}{\frac {e^{x}+e^{-x}}{2}}dx\\&={\frac {e^{x}-e^{-x}}{2}}{\bigg |}_{0}^{1}\\&=\mathbf {\frac {e-{\frac {1}{e}}}{2}} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c37fa7e5bde769230d12416141f2e0b9080ed7f)
3. Find the circumference of the circle given by the parametric equations
![{\displaystyle x(t)=R\cos(t),y(t)=R\sin(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b680b5d879f80be573f553cef5ded0b9d62c297)
, with
![{\displaystyle t\in [0,2\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dbc9ed8510c75442ce1d2e73f021258fc7e04c6)
.
:
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{2\pi }{\sqrt {\left({\tfrac {d}{dt}}{\big (}R\cos(t){\big )}\right)^{2}+\left({\tfrac {d}{dt}}{\big (}R\sin(t){\big )}\right)^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {{\big (}-R\sin(t){\big )}^{2}+{\big (}R\cos(t){\big )}^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {R^{2}{\big (}\sin ^{2}(t)+\cos ^{2}(t){\big )}}}dt\\&=\int \limits _{0}^{2\pi }Rdt\\&=R\cdot t{\Big |}_{0}^{2\pi }\\&=\mathbf {2\pi R} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb61ab3a36536f1b8ef4b6e557dda44b7a2349d7)
:
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{2\pi }{\sqrt {\left({\tfrac {d}{dt}}{\big (}R\cos(t){\big )}\right)^{2}+\left({\tfrac {d}{dt}}{\big (}R\sin(t){\big )}\right)^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {{\big (}-R\sin(t){\big )}^{2}+{\big (}R\cos(t){\big )}^{2}}}dt\\&=\int \limits _{0}^{2\pi }{\sqrt {R^{2}{\big (}\sin ^{2}(t)+\cos ^{2}(t){\big )}}}dt\\&=\int \limits _{0}^{2\pi }Rdt\\&=R\cdot t{\Big |}_{0}^{2\pi }\\&=\mathbf {2\pi R} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb61ab3a36536f1b8ef4b6e557dda44b7a2349d7)
4. Find the length of one arch of the cycloid given by the parametric equations
![{\displaystyle x(t)=R{\big (}t-\sin(t){\big )},y(t)=R{\big (}1-\cos(t){\big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/147added64964c2e36399d5bb8b8e2c5647708aa)
, with
![{\displaystyle t\in [0,2\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dbc9ed8510c75442ce1d2e73f021258fc7e04c6)
.
:
Using the trigonometric identity
![{\displaystyle \sin ^{2}\left({\frac {t}{2}}\right)={\frac {1-\cos(t)}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49f060af33025ffa2032441fa968fcb799bad8bc)
, we have
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{2\pi }R{\sqrt {4\sin ^{2}\left({\frac {t}{2}}\right)}}dt\\&=\int \limits _{0}^{2\pi }2R\sin \left({\frac {t}{2}}\right)dt\\&=-4R\cos \left({\frac {t}{2}}\right){\Bigg |}_{0}^{2\pi }\\&=-4R(-1-1)\\&=\mathbf {8R} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/145149e56b3e05651b3b1ed3dbc7ea17686467fc)
:
Using the trigonometric identity
![{\displaystyle \sin ^{2}\left({\frac {t}{2}}\right)={\frac {1-\cos(t)}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49f060af33025ffa2032441fa968fcb799bad8bc)
, we have
![{\displaystyle {\begin{aligned}s&=\int \limits _{0}^{2\pi }R{\sqrt {4\sin ^{2}\left({\frac {t}{2}}\right)}}dt\\&=\int \limits _{0}^{2\pi }2R\sin \left({\frac {t}{2}}\right)dt\\&=-4R\cos \left({\frac {t}{2}}\right){\Bigg |}_{0}^{2\pi }\\&=-4R(-1-1)\\&=\mathbf {8R} \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/145149e56b3e05651b3b1ed3dbc7ea17686467fc)