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Net displacement via integration of velocity

A robot arm moves along a straight track. Its velocity at time tt (in seconds) is given by v(t)=ln⁡(t+1)v(t) = \ln(t+1) meters per second, valid for t≥0t \geq 0.

Using the Net Change Theorem, find the total (net) displacement of the robot arm over the time interval [0,e−1][0, e-1]. That is, compute

∫0e−1ln⁡(t+1) dt\int_0^{e-1} \ln(t+1)\, dt

What is the robot arm's net displacement, in meters?

Hint

Let u=t+1u = t+1, so du=dtdu = dt. This turns the integral into ∫ln⁡u du\int \ln u \, du, which requires integration by parts: ∫ln⁡u du=uln⁡u−u+C\int \ln u\, du = u\ln u - u + C. Don't forget to transform the bounds: when t=0t=0, u=1u=1; when t=e−1t=e-1, u=eu=e.

Step-by-step solution

Substitute u=t+1u = t+1, so du=dtdu = dt. When t=0t=0, u=1u=1, and when t=e−1t = e-1, u=eu = e. The integral becomes ∫1eln⁡u du.\int_1^e \ln u\, du.

Using integration by parts (or the known antiderivative), ∫ln⁡u du=uln⁡u−u+C\int \ln u\, du = u\ln u - u + C. Evaluating from 11 to ee: at u=eu=e, we get eln⁡e−e=e(1)−e=0e\ln e - e = e(1) - e = 0; at u=1u=1, we get 1ln⁡1−1=0−1=−11\ln 1 - 1 = 0 - 1 = -1. So the result is 0−(−1)=10 - (-1) = 1.

Therefore the net displacement is 11 meter — even though velocity is always nonnegative on this interval, the small positive area near t=0t=0 balances precisely to give exactly 11 m of net displacement.

Sign of velocity on the interval

Since velocity was never negative, was total distance traveled equal to net displacement?

Hint
Step-by-step solution

Yes, total distance traveled equals net displacement in this case. Since velocity, the natural log of t plus one, is nonnegative for all t greater than or equal to zero, the arm never moved backward. Because there is no sign change, splitting the integral at zeros of velocity is unnecessary, and the integral of the absolute value of velocity equals the plain integral of velocity, which was found to be one meter.

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