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- anonymous

The force F (in newtons) of a hydraulic cylinder in a press is proportional to the square of sec x where x is the distance (in meters) that the cylinder is extended in its cycle. The domain of F is [0, π/3], and F(0) = 900
Find the average force exerted by the press over the interval [0, π/3]. (Round your answer to one decimal place.)

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- anonymous

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- marco26

Perhaps you can use: \[F _{ave}=\frac{ f(x _{final})-f(x _{initial}){} }{ x _{final}-x _{initial} }\]

- IrishBoy123

we have \(F(x) = k \sec^2 x\)
and \(F(0) = 900 \implies k = 900\)
so you want the average value of \(F(x) = 900 \sec^2 x\) over the interval \(0 \le x \le \frac{\pi}{3}\)
|dw:1443517655585:dw|
for the average, geometrically you want to find the area under the \(F=900\sec^2(x)\) curve in the interval and then find the rectangle with equivalent area. the height of that rectangle is the average value of F through the cycle
in physical terms, the work done by the piston is the area under the \(F=900\sec^2(x)\) curve, ie \(W=\int F dx\). the average value of F is the constant force F that would do the same amount of work: \(W=\int F dx = \bar F\times x\).
in either case you can say average force \(\bar F \) is
\[\large \bar F = \dfrac{\int\limits\limits_{0}^{\frac{\pi}{3}} F(x) \; dx}{\frac{\pi}{3}} = \dfrac{900\int\limits_{0}^{\frac{\pi}{3}} \sec^2x \; dx}{\frac{\pi}{3}} \]
[NB typo in drawing, it's \(sec^\color{red}2 x\)]

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