# How do you find the definite integral for: tln(t) dt for the intervals [1, 10]?

##### 1 Answer
Jul 13, 2016

$= 25 \left(\ln 100 - 1\right)$

#### Explanation:

${\int}_{0}^{10} \setminus t \ln t \setminus \mathrm{dt}$

$= {\int}_{0}^{10} \setminus \frac{d}{\mathrm{dt}} \left({t}^{2} / 2\right) \ln t \setminus \mathrm{dt}$

and by IBP: $\int u v ' = u v - \int u ' v$

$= {t}^{2} / 2 \ln t {|}_{0}^{10} - {\int}_{0}^{10} \setminus {t}^{2} / 2 \frac{d}{\mathrm{dt}} \left(\ln t\right) \setminus \mathrm{dt}$

$= {t}^{2} / 2 \ln t {|}_{0}^{10} - {\int}_{0}^{10} \setminus {t}^{2} / 2 \frac{1}{t} \setminus \mathrm{dt}$

$= {t}^{2} / 2 \ln t {|}_{0}^{10} - {\int}_{0}^{10} \setminus \frac{t}{2} \setminus \mathrm{dt}$

$= {t}^{2} / 2 \ln t - {t}^{2} / 4 {|}_{0}^{10} q \quad \triangle$

clearly ${t}^{2} / 2 \ln t {|}_{t = 0}$ is a problem and we need to know

${\lim}_{t \to {0}^{+}} {t}^{2} / 2 \ln t$

$= \frac{1}{2} {\lim}_{t \to {0}^{+}} \ln \frac{t}{\frac{1}{t} ^ 2}$ which is now indeterminate so L'Hopital'sRule

$= \frac{1}{2} {\lim}_{t \to {0}^{+}} \frac{\frac{1}{t}}{- \frac{2}{t} ^ 3}$

$= - \frac{1}{4} {\lim}_{t \to {0}^{+}} {t}^{2} = 0$

so $\triangle$ becomes

$= {10}^{2} / 2 \ln 10 - {10}^{2} / 4$

$= {10}^{2} / 4 \cdot 2 \ln 10 - {10}^{2} / 4$

$= {10}^{2} / 4 \left(\ln 100 - 1\right)$

$= 25 \left(\ln 100 - 1\right)$