integral from-1 to 1 of (2e^x-2)-(x^2-5)

解

\int_{- 1}^{1} (2 e^{x} - 2) - (x^{2} - 5) d x

2 (e - \frac{1}{e}) + \frac{16}{3}

十進法表記

10.03413 \dots

解答ステップ

\int_{- 1}^{1} 2 e^{x} - 2 - (x^{2} - 5) 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}^{1} 2 e^{x} d x - \int_{- 1}^{1} 2 d x - \int_{- 1}^{1} x^{2} - 5 d x

\int_{- 1}^{1} 2 e^{x} d x = 2 (e - \frac{1}{e})

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

\int_{- 1}^{1} x^{2} - 5 d x = - \frac{28}{3}

= 2 (e - \frac{1}{e}) - 4 - (- \frac{28}{3})

簡素化

2 (e - \frac{1}{e}) - 4 - (- \frac{28}{3}) : 2 (e - \frac{1}{e}) + \frac{16}{3}

= 2 (e - \frac{1}{e}) + \frac{16}{3}