Answer :
[tex]\int \limits_2^5x^2+3x\, dx=\\
\Big[\frac{x^3}{3}+\frac{3x^2}{2}\Big]_2^5=\\
\frac{5^3}{3}+\frac{3\cdot5^2}{2}-(\frac{2^3}{3}+\frac{3\cdot2^2}{2})=\\
\frac{125}{3}+\frac{75}{2}-(\frac{8}{3}+6)=\\
\frac{250}{6}+\frac{225}{6}-(\frac{16}{6}+\frac{36}{6})=\\
\frac{475}{6}-\frac{52}{6}=\\
\frac{423}{6}=\frac{141}{2}=70.5[/tex]
