integral from-3 to 4 of x^2+sin(x)+e^x

解

\int_{- 3}^{4} x^{2} + sin (x) + e^{x} d x

\frac{91}{3} - cos (4) + cos (3) + e^{4} - \frac{1}{e ^{3}}

十進法表記

84.54534 \dots

解答ステップ

\int_{- 3}^{4} x^{2} + sin (x) + e^{x} 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_{- 3}^{4} x^{2} d x + \int_{- 3}^{4} sin (x) d x + \int_{- 3}^{4} e^{x} d x

\int_{- 3}^{4} x^{2} d x = \frac{91}{3}

\int_{- 3}^{4} sin (x) d x = - cos (4) + cos (3)

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

= \frac{91}{3} - cos (4) + cos (3) + e^{4} - \frac{1}{e ^{3}}