integral from-1 to 0 of (2x+3)^2

解

\int_{- 1}^{0} (2 x + 3)^{2} d x

\frac{13}{3}

十進法表記

4.33333 \dots

解答ステップ

\int_{- 1}^{0} (2 x + 3)^{2} d x

拡張

(2 x + 3)^{2} : 4 x^{2} + 12 x + 9

= \int_{- 1}^{0} 4 x^{2} + 12 x + 9 d x

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}^{0} 4 x^{2} d x + \int_{- 1}^{0} 12 x d x + \int_{- 1}^{0} 9 d x

\int_{- 1}^{0} 4 x^{2} d x = \frac{4}{3}

\int_{- 1}^{0} 12 x d x = - 6

\int_{- 1}^{0} 9 d x = 9

= \frac{4}{3} - 6 + 9

簡素化

\frac{4}{3} - 6 + 9 : \frac{13}{3}

= \frac{13}{3}