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Find the total area between the curve f(x)=cosx and the x-axis on the interval[0,2pi].

Here, we have to be very careful. The question asks for the total area between the function and the x-axis but if we simply take the integral,

\int _{{0}}^{{2\pi }}\cos x\,dx=\sin x{\bigg |}_{{0}}^{{2\pi }}=0

notice that we get an answer of 0. This is because the integral counts area above the x-axis as positive and that below the x-axis as negative. This problem contains an equal amount of area above and below the graph. So, we need to split this up into a few integrals

A=\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos x\,dx-\int _{{{\frac  {\pi }{2}}}}^{{{\frac  {3\pi }{2}}}}\cos x\,dx+\int _{{{\frac  {3\pi }{2}}}}^{{2\pi }}\cos x\,dx\,

Notice the middle integral has a negative sign out in front. This is because the graph is beneath the x-axis on that interval. Since the integral counts it as negative area, a negative sign in front makes it positive. The two integrals without negative signs in front represent regions where the function is above the x-axis. So, the area is

A=\sin x{\bigg |}_{{0}}^{{{\frac  {\pi }{2}}}}-\sin x{\bigg |}_{{{\frac  {\pi }{2}}}}^{{{\frac  {3\pi }{2}}}}+\sin x{\bigg |}_{{{\frac  {3\pi }{2}}}}^{{2\pi }}=(1-0)-(-1-1)+(0--1)=4\,

Thus the total area between the curve f(x)=\cos x and the x-axis on the interval [0,2\pi ] is 4 square units.

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