Question

$\underset{1}{\overset{2}{\int }}x\sqrt{3x-2}dx$

Answer 1

$$\begin{gathered} {\int }_1^{2}x\sqrt{3x-2}dx \\ \mathrm{Let},3x-2=t,\mathrm{then}3dx=dt \\ \mathrm{when},x=1;t=1\mathrm{and}x=2;t=4 \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ {\int }_1^4\frac{t+2}{3}\sqrt{t}\frac{dt}{3} \\ =\frac{1}{9}{\int }_1^4{t}^{\frac{3}{2}}+2\sqrt{t}dt \\ =\frac{1}{9}{\left[\frac{2t^{{\displaystyle \frac{5}{2}}}}{5}+\frac{4t^{{\displaystyle \frac{3}{2}}}}{3}\right]}_1^4 \\ =\frac{1}{9}\left[\frac{64}{5}+\frac{32}{3}-\frac{2}{5}-\frac{4}{3}\right] \\ =\frac{46}{135} \end{gathered}$$