Find the numerical value of each integral function with the given condition:
500 = initial amount in tons G(s) is amount arriving in tons per hour and 100 is amount processed in tons per hour. G(s)-100 is the rate at which the amount of gravel is changing in tons per hour. The integral is essentially (tons/hour)(hours) = tons of gravel, and therefore gives the net/accumulated amount of unprocessed gravel at the plant at any time t (which gets added to the 500 tons already present)
tA(t) 00 4.92348635.376 8525.551 The maximum amount of unprocessed gravel during the given workday is 635.376 tons.