$\text { The force } F \text { (in newtons) of a hydraulic cylinder in a press is proportional to the }$ square of $\sec x$ where $x$ is the distance (in meters) that the cylinder is extended in its cycle. The domain of $F$ is $[ 0 , \pi / 3 ]$ and $F ( 0 ) = 400$. Find the average force exerted by the press over the interval $[ 0 , \pi / 3 ]$. Round your answer to the nearest whole number.
A) 221 newtons
B) 110 newtons
C) 540 newtons
D) 180 newtons
E) 662 newtons

Question

Quizplus · Accepted Answer

The force $ F $ is proportional to $ \sec^2 x $, so $ F(x) = k \sec^2 x $. Given $ F(0) = 400 $, we find $ k = 400 $. The average force is $\frac{1}{\pi/3 - 0} \int_0^{\pi/3} 400 \sec^2 x \, dx = \frac{3}{\pi} [400 \tan x]_0^{\pi/3} = \frac{3}{\pi} (400 \sqrt{3}) = 662$ newtons.

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