integral from-1 to 1 of e^x-x^2+6

Lösung

\int_{- 1}^{1} e^{x} - x^{2} + 6 d x

e + \frac{34}{3} - \frac{1}{e}

Dezimale

13.68373 \dots

Schritte zur Lösung

\int_{- 1}^{1} e^{x} - x^{2} + 6 d x

Wende die Summenregel an:

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

= \int_{- 1}^{1} e^{x} d x - \int_{- 1}^{1} x^{2} d x + \int_{- 1}^{1} 6 d x

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

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

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

= e - \frac{1}{e} - \frac{2}{3} + 12

Ziehe Brüche zusammen

12 - \frac{2}{3} : \frac{34}{3}

= e + \frac{34}{3} - \frac{1}{e}

x \to 5 lim (\frac{2 - 25}{x - 5})

n = 1 \sum \infty \frac{1}{e ^{n + 1}}

s l o p e (3, - 4), (5, 1)

\int \frac{x + 2}{x ^{2} - 1} d x

d er i v a t i v e cos (2 θ)