Consider the following problem: The depth of the snow on Cam's driveway is increasing at a rate of r(t)=(t+1)/(2) centimeters per hour (where t is the time in hours). At time t=5, the depth of the snow is 9 centimeters. By how much does the depth of the snow increase between hours 5 and 8 ? Which expression can we use to solve the problem? Choose 1 answer: (A) int_(5)^(8)r(t)dt (B) 9+int_(5)^(8)r(t)dt (C) r^(')(8)-9 (D) r^(')(8)

Question

Consider the following problem: The depth of the snow on Cam's driveway is increasing at a rate of  r(t)=(t+1)/(2) centimeters per hour (where  t is the time in hours). At time  t=5, the depth of the snow is 9 centimeters. By how much does the depth of the snow increase between hours 5 and 8 ? Which expression can we use to solve the problem? Choose 1 answer: (A)  int_(5)^(8)r(t)dt (B)  9+int_(5)^(8)r(t)dt (C)  r^(')(8)-9 (D)  r^(')(8)

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Integrate rate of increase: To find the increase in the depth of the snow between hours 5 and 8, we need to integrate the rate of increase, r(t), from t=5 to t=8. This will give us the total change in depth over that time period.Use correct expression: The correct expression to use for this problem is the integral of r(t) from 5 to 8, which is represented by the integral sign with the limits of integration. This corresponds to option (A) int_(5)^(8)r(t)dt.Calculate integral of r(t): We will now calculate the integral of r(t) from 5 to 8. The function r(t) is given by (t+1)/(2). The integral of r(t) with respect to t is the antiderivative of (t+1)/(2) evaluated from t=5 to t=8.Evaluate antiderivative at t=8: The antiderivative of (t+1)/(2) is (t^2/2 + t)/2. We will evaluate this antiderivative at t=8 and t=5 and then subtract the value at t=5 from the value at t=8 to find the total change in depth.Evaluate antiderivative at t=5: Evaluating the antiderivative at t=8 gives ((8^2)/2 + 8)/2 = (64/2 + 8)/2 = (32 + 8)/2 = 40/2 = 20 centimeters.Find total change in depth: Evaluating the antiderivative at t=5 gives ((5^2)/2 + 5)/2 = (25/2 + 5)/2 = (12.5 + 5)/2 = 17.5/2 = 8.75 centimeters.The increase in depth from t=5 to t=8 is the difference between the antiderivative evaluated at t=8 and t=5, which is 20 cm - 8.75 cm = 11.25 centimeters.

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