$$\int_1^2 \frac{(x+3)^2}{3} dx$$
solve this
Hello Guest!
\({\color{blue}\int_1^2 \frac{(x+3)^2}{3}\ dx}=\int_1^2\frac{1}{3}(x^2+6x+9)\ dx=\frac{1}{3}\left[\frac{x^3}{3}+3x^2+9x\right]_{1}^{2}\\ = \frac{1}{3}(\frac{8}{3}+12+18-\frac{1}{3}-3-9)\color{blue}=6\frac{7}{9}\)
!
!
