integral from-1 to 1/2 of (t+1)(t^2+4)

解

\int_{- 1}^{\frac{1}{2}} (t + 1) (t^{2} + 4) d t

\frac{297}{64}

十進法表記

4.640625

解答ステップ

\int_{- 1}^{\frac{1}{2}} (t + 1) (t^{2} + 4) d t

拡張

(t + 1) (t^{2} + 4) : t^{3} + 4 t + t^{2} + 4

= \int_{- 1}^{\frac{1}{2}} t^{3} + 4 t + t^{2} + 4 d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{\frac{1}{2}} t^{3} d t + \int_{- 1}^{\frac{1}{2}} 4 t d t + \int_{- 1}^{\frac{1}{2}} t^{2} d t + \int_{- 1}^{\frac{1}{2}} 4 d t

\int_{- 1}^{\frac{1}{2}} t^{3} d t = - \frac{15}{64}

\int_{- 1}^{\frac{1}{2}} 4 t d t = - \frac{3}{2}

\int_{- 1}^{\frac{1}{2}} t^{2} d t = \frac{3}{8}

\int_{- 1}^{\frac{1}{2}} 4 d t = 6

= - \frac{15}{64} - \frac{3}{2} + \frac{3}{8} + 6

簡素化

- \frac{15}{64} - \frac{3}{2} + \frac{3}{8} + 6 : \frac{297}{64}

= \frac{297}{64}