∫ 1 4 t ( 1 + t ) d t = ∫ 1 4 ( t 1 2 + t 3 2 ) d t = ( 2 t 3 2 3 + 2 t 5 2 5 ) | 1 4 = [ 2 ( 4 ) 3 2 3 + 2 ( 4 ) 5 2 5 ] − [ 2 ( 1 ) 3 2 3 + 2 ( 1 ) 5 2 5 ] = [ 16 3 + 64 5 ] − [ 2 3 + 2 5 ] = 256 15 {\displaystyle {\begin{aligned}\int _{1}^{4}{\sqrt {t}}(1+t)dt&=\int _{1}^{4}\left(t^{\frac {1}{2}}+t^{\frac {3}{2}}\right)dt\\[2ex]&=\left({\frac {2t^{\frac {3}{2}}}{3}}+{\frac {2t^{\frac {5}{2}}}{5}}\right){\Bigg |}_{1}^{4}\\[2ex]&=\left[{\frac {2(4)^{\frac {3}{2}}}{3}}+{\frac {2(4)^{\frac {5}{2}}}{5}}\right]-\left[{\frac {2(1)^{\frac {3}{2}}}{3}}+{\frac {2(1)^{\frac {5}{2}}}{5}}\right]\\[2ex]&=\left[{\frac {16}{3}}+{\frac {64}{5}}\right]-\left[{\frac {2}{3}}+{\frac {2}{5}}\right]\\[2ex]&={\frac {256}{15}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{4}{\sqrt {t}}(1+t)dt&=\int _{1}^{4}\left(t^{\frac {1}{2}}+t^{\frac {3}{2}}\right)dt\\[2ex]&=\left({\frac {2t^{\frac {3}{2}}}{3}}+{\frac {2t^{\frac {5}{2}}}{5}}\right){\Bigg |}_{1}^{4}\\[2ex]&=\left[{\frac {2(4)^{\frac {3}{2}}}{3}}+{\frac {2(4)^{\frac {5}{2}}}{5}}\right]-\left[{\frac {2(1)^{\frac {3}{2}}}{3}}+{\frac {2(1)^{\frac {5}{2}}}{5}}\right]\\[2ex]&=\left[{\frac {16}{3}}+{\frac {64}{5}}\right]-\left[{\frac {2}{3}}+{\frac {2}{5}}\right]\\[2ex]&={\frac {256}{15}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/fda02451a09d311d14ed3b2910293328b5ffe18a)