∫ 0 2 x ( 2 + x 5 ) d x = ∫ 0 2 ( 2 x + x 6 ) d x = ∫ 0 2 ( 2 x + x 6 ) d x = ( 2 x 1 + 1 1 + 1 + x 6 + 1 6 + 1 ) | 0 2 = ( x 2 + x 7 7 ) | 0 2 = ( ( 2 ) 2 − ( 2 ) 7 7 ) − ( ( 0 ) 2 + ( 0 ) 7 7 ) = [ 4 + 2 7 7 ] − [ 0 ] = 156 7 {\displaystyle {\begin{aligned}\int _{0}^{2}x(2+x^{5})\,dx&=\int _{0}^{2}(2x+x^{6})\,dx=\int _{0}^{2}(2x+x^{6})\,dx\\[2ex]&=\left({\frac {2x^{1+1}}{1+1}}+{\frac {x^{6+1}}{6+1}}\right){\bigg |}_{0}^{2}=\left(x^{2}+{\frac {x^{7}}{7}}\right){\bigg |}_{0}^{2}\\[2ex]&=\left((2)^{2}-{\frac {(2)^{7}}{7}}\right)-\left((0)^{2}+{\frac {(0)^{7}}{7}}\right)\\[2ex]&=\left[4+{\frac {2^{7}}{7}}\right]-[0]\\[2ex]&={\frac {156}{7}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{0}^{2}x(2+x^{5})\,dx&=\int _{0}^{2}(2x+x^{6})\,dx=\int _{0}^{2}(2x+x^{6})\,dx\\[2ex]&=\left({\frac {2x^{1+1}}{1+1}}+{\frac {x^{6+1}}{6+1}}\right){\bigg |}_{0}^{2}=\left(x^{2}+{\frac {x^{7}}{7}}\right){\bigg |}_{0}^{2}\\[2ex]&=\left((2)^{2}-{\frac {(2)^{7}}{7}}\right)-\left((0)^{2}+{\frac {(0)^{7}}{7}}\right)\\[2ex]&=\left[4+{\frac {2^{7}}{7}}\right]-[0]\\[2ex]&={\frac {156}{7}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/62167bee792832f664296ca9420fc370b9be2361)