# How do you evaluate the definite integral int sqrtt ln(t)dt from 2 to 1?

Jul 14, 2016

$\frac{4}{9} \left(2 \sqrt{2} - 1\right) - \left(\frac{4 \sqrt{2}}{3}\right) \ln 2.$

#### Explanation:

We use the Fundamental Theorem of Integral Calculus

: intf(t)dt=F(t) rArr ""int_a^bf(t)dt=F(b)-F(a).

So, we first find out the Indefinite Integral $I = \int \sqrt{t} \ln t \mathrm{dt} ,$ using the

Method of Integration by Parts,which states that,

$\int u v \mathrm{dt} = u \int v \mathrm{dt} - \int \left\{\frac{\mathrm{du}}{\mathrm{dt}} \int v \mathrm{dt}\right\} \mathrm{dt} .$

We apply this by taking u=lnt, &, v=sqrtt, so that,

(du)/dt=1/t, &, intvdt=(t^(3/2))/(3/2)=(2/3)t^(3/2).

$\therefore I = \left(\frac{2}{3}\right) {t}^{\frac{3}{2}} \ln t - \int \left\{\frac{1}{t} \cdot \left(\frac{2}{3}\right) {t}^{\frac{3}{2}}\right\} \mathrm{dt} ,$

$= \left(\frac{2}{3}\right) {t}^{\frac{3}{2}} \ln t - \left(\frac{2}{3}\right) \int {t}^{\frac{1}{2}} \mathrm{dt} ,$

$= \left(\frac{2}{3}\right) {t}^{\frac{3}{2}} \ln t - \left(\frac{2}{3}\right) \left(\frac{2}{3}\right) {t}^{\frac{3}{2}}$,

$\therefore I = \left(\frac{2}{3}\right) {t}^{\frac{3}{2}} \ln t - \left(\frac{4}{9}\right) {t}^{\frac{3}{2}} + C .$

Hence, ${\int}_{2}^{1} \int \sqrt{t} \ln t \mathrm{dt} = \left[\left(\frac{2}{3}\right) \cdot {1}^{\frac{3}{2}} \cdot \ln 1 - \left(\frac{4}{9}\right) \cdot {1}^{\frac{3}{2}}\right] - \left[\left(\frac{2}{3}\right) \cdot {2}^{\frac{3}{2}} \cdot \ln 2 - \left(\frac{4}{9}\right) \cdot {2}^{\frac{3}{2}}\right]$

$= \left[0 - \frac{4}{9}\right] - \left[\frac{{2}^{\frac{5}{2}} \cdot \ln 2}{3} - \frac{4}{9} \cdot {2}^{\frac{3}{2}}\right]$

$= \frac{4}{9} \left({2}^{\frac{3}{2}} - 1\right) - {2}^{\frac{5}{2}} / 3 \cdot \left(\ln 2\right)$

$= \frac{4}{9} \left(2 \sqrt{2} - 1\right) - \left(\frac{4}{3} \sqrt{2}\right) \ln 2.$

Jul 14, 2016

$= \frac{2}{3} \left(\frac{4 \sqrt{2}}{3} - \frac{2}{3} - 2 \sqrt{2} \cdot \ln 2\right)$

#### Explanation:

${\int}_{2}^{1} \sqrt{t} \ln \left(t\right) \mathrm{dt}$

$= {\int}_{2}^{1} \frac{d}{\mathrm{dt}} \left(\frac{2}{3} {t}^{\frac{3}{2}}\right) \ln \left(t\right) \mathrm{dt}$

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

$= {\left[\frac{2}{3} {t}^{\frac{3}{2}} \cdot \ln \left(t\right)\right]}_{2}^{1} - {\int}_{2}^{1} \frac{2}{3} {t}^{\frac{3}{2}} \setminus \frac{d}{\mathrm{dt}} \left(\ln \left(t\right)\right) \setminus \mathrm{dt}$

$= {\left[\frac{2}{3} {t}^{\frac{3}{2}} \cdot \ln \left(t\right)\right]}_{2}^{1} - {\int}_{2}^{1} \frac{2}{3} {t}^{\frac{3}{2}} \setminus \frac{1}{t} \setminus \mathrm{dt}$

$= {\left[\frac{2}{3} {t}^{\frac{3}{2}} \cdot \ln \left(t\right)\right]}_{2}^{1} - \frac{2}{3} {\int}_{2}^{1} {t}^{\frac{1}{2}} \setminus \mathrm{dt}$

$= \frac{2}{3} {\left[{t}^{\frac{3}{2}} \cdot \ln \left(t\right) - \frac{2}{3} {t}^{\frac{3}{2}} \setminus\right]}_{2}^{1}$

$= \frac{2}{3} {\left[{t}^{\frac{3}{2}} \left(\ln \left(t\right) - \frac{2}{3}\right) \setminus\right]}_{2}^{1}$

$= \frac{2}{3} \left(\left[1 \cdot \left(0 - \frac{2}{3}\right) \setminus\right] - \left[2 \sqrt{2} \cdot \left(\ln 2 - \frac{2}{3}\right) \setminus\right]\right)$

$= \frac{2}{3} \left(\frac{4 \sqrt{2}}{3} - \frac{2}{3} - 2 \sqrt{2} \cdot \ln 2\right)$