anonymous
  • 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.)
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
marco26
  • marco26
Perhaps you can use: \[F _{ave}=\frac{ f(x _{final})-f(x _{initial}){} }{ x _{final}-x _{initial} }\]
IrishBoy123
  • 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 \(sec^\color{red}2 x\)]

Looking for something else?

Not the answer you are looking for? Search for more explanations.