I've been struggling with integration by parts. Can anyone explain these problems to me?
Help is very much appreciated. Thank you!
Recall that
$\displaystyle \int{u\,dv} = uv - \int{v\,du}$.
For Q.1 let $\displaystyle u = \ln{t}$ so that $\displaystyle du = \frac{1}{t}$, and let $\displaystyle dv = 7t^2$ so that $\displaystyle v = \frac{7}{3}t^3$.
Thus
$\displaystyle \int{7t^2 \ln{t}\,dt} = \frac{7}{3}t^3\ln{t} - \int{\frac{7}{3}t^3\,\frac{1}{t}\,dt}$
$\displaystyle = \frac{7}{3}t^3\ln{t} - \frac{7}{3}\int{t^2\,dt}$
$\displaystyle = \frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3 + C$.
Can you now find $\displaystyle \int_1^e{7t^2\ln{t}\,dt}$?
Q.1
$\displaystyle \int_1^e{7t^2\ln{t}\,dt} = \left[\frac{7}{3}t^3\ln{t} - \frac{7}{9}t^3\right]_1^e$
$\displaystyle = \left[\frac{7}{3}e^3\ln{e} - \frac{7}{9}e^3\right] - \left[\frac{7}{3}(1)^3 \ln{1} - \frac{7}{9}(1)^3\right]$
$\displaystyle = \left[\frac{7}{3}e^3 - \frac{7}{9}e^3\right] - \left[- \frac{7}{9}\right]$
$\displaystyle = \frac{14}{9}e^3 + \frac{7}{9}$.
Q.2
$\displaystyle \int_0^6{t e^{-t}\,dt} = \left[-te^{-t} - e^{-t}\right]_0^6$
$\displaystyle = \left[-6e^{-6} - e^{-6}\right] - \left[-0e^{-0} - e^{-0}\right]$
$\displaystyle = -7e^{-6} - 1$.