integral from 0 to 1 of 6x^2+4x-3+4e^{4x}

解

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 x} d x

e^{4}

十進法表記

54.59815 \dots

解答ステップ

\int_{0}^{1} 6 x^{2} + 4 x - 3 + 4 e^{4 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_{0}^{1} 6 x^{2} d x + \int_{0}^{1} 4 x d x - \int_{0}^{1} 3 d x + \int_{0}^{1} 4 e^{4 x} d x

\int_{0}^{1} 6 x^{2} d x = 2

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

\int_{0}^{1} 3 d x = 3

\int_{0}^{1} 4 e^{4 x} d x = e^{4} - 1

= 2 + 2 - 3 + e^{4} - 1

簡素化

= e^{4}