## Calculus 8th Edition

$2-e^{-t}(t^2+2t+2)$ meters
Integration by parts formula: $\displaystyle \int fg'$ $\displaystyle dx = fg-\int f'g$ $dx$ Distance $= \displaystyle \int_0^t|v(t)|dt\quad$but since $v(t)$ > 0 for $t > 0$ we can drop the absolute value $\displaystyle \int_0^tv(t)dt = \displaystyle \int_0^tt^2e^{-t}dt$ ____________________________ Let $\displaystyle f = t^2$ and $\displaystyle g' = e^{-t}$ thus $\displaystyle f' = 2t$ and $\displaystyle g = -e^{-t}$ ____________________________ $\displaystyle \int_0^tt^2e^{-t}dt = [t^2(-e^{-t})]_0^t-\int_0^t(2t)(-e^{-t})dt$ $\displaystyle [t^2(-e^{-t}) - ((0)^2(-e^0))] + 2\int_0^tte^{-t}dt$ $\displaystyle [-t^2e^{-t} -0]+ 2[(t[-e^{-t}])|_0^t-\int_0^t-e^{-t}(1)dt]$ $\displaystyle -t^2e^{-t} + 2[(-te^{-t})|_0^t+\int_0^te^{-t}dt]$ $\displaystyle -t^2e^{-t} + 2([-te^{-t} - (-(0)e^0)] + [-e^{-t}]_0^t)$ $\displaystyle -t^2e^{-t} + 2([-te^{-t} + 0]+[-e^{-t} - (-e^0)])$ $\displaystyle -t^2e^{-t} + 2(-te^{-t}-e^{-t} - (-1))$ $\displaystyle -t^2e^{-t} + -2te^{-t}-2e^{-t} + 2(1)$ $2-e^{-t}(t^2+2t+2)$ meters