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Calculus again!

A yo-yo is moving up and down a string so that its velocity at time t is given by v(t) = 4cos(t) for time t ≥ 0. The initial position of the yo-yo at time t = 0 is x = 4. Find the displacement and distance from time t = 0 to time t = π.
I know that the displacement would be calculated by [tex]\int\limits^0_\pi {4cos(t)} \, dx[/tex] and the displacement would be calculated by [tex]\int\limits^0_\pi {|4cos(t)|} \, dx[/tex], but would both be 0? Thank you!

Sagot :

Not. Both are diferent.

The displacement  from time t = 0 to time t = π is given by:

[tex]{\displaystyle s = \int _0^{\pi }4\cos\left(t\right)\:dt}[/tex]

[tex]{\displaystyle s = 4\left[\sin \left(t\right)\right]^{\pi }_0}[/tex]

[tex]s = 4[0 - 0 ][/tex]

[tex]s = 0[/tex]

The distance from time t = 0 to time t = π is:

[tex]{\displaystyle d = \int _0^{\pi }|4\cos\left(t\right)|\:dt}[/tex]

[tex]{\displaystyle d = \int _0^{\frac{\pi }{2}}4\cos \left(t\right)dt+\int _{\frac{\pi }{2}}^{\pi }-4\cos \left(t\right)dt}[/tex]

[tex]d = 4\left[\sin \left(t\right)\right]^{\frac{\pi}{2}}_0 - 4\left[\sin \left(t\right)\right]^{\pi}_{\frac{\pi}{2}}[/tex]

[tex]d = 4 + 4[/tex]

[tex]d = 8[/tex]