The velocity of a point's motion is v=te^(-t/100). Find

The velocity of a point's motion is $v=te^{-t/100}$. Find the length of the path traveled by the point from the beginning to the end of the motion.

Step 1: Set up the improper integral

\[s=\int_0^{\infty}te^{-t/100}\,dt.\]

Step 2: Antiderivative by parts

With $u=t,\ dv=e^{-t/100}dt$, so $du=dt,\ v=-100e^{-t/100}$:

\[\int te^{-t/100}\,dt=-100te^{-t/100}+100\int e^{-t/100}\,dt=-100e^{-t/100}(t+100)+C.\]

Step 3: Evaluate the limits

As $t\to\infty$, $-100e^{-t/100}(t+100)\to0$ (exponential decay dominates the linear factor, by L'Hôpital's rule). At $t=0$: $-100\cdot1\cdot100=-10000$.

\[s=0-(-10000)=10000.\]

\[\boxed{s=10000}\]

Step 4: Cross-check against the standard formula

The general result $\int_0^\infty te^{-\alpha t}\,dt=1/\alpha^2$ with $\alpha=1/100$ gives $s=100^2=10000$, confirming the computation.

Step 5: Physical picture

The velocity rises from $0$, peaks at $t=100$ (where $dv/dt=0$) with $v_{\max}=100/e\approx36.8$, and then decays toward zero; because the decay is exponential rather than polynomial, the area under the whole infinite curve — the total distance — stays finite.