integral from 4 to 1 of 4x+3x^2-2x^3

解

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

\frac{69}{2}

十進法表記

34.5

解答ステップ

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{4} 4 x + 3 x^{2} - 2 x^{3} 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}^{4} 4 x d x + \int_{1}^{4} 3 x^{2} d x - \int_{1}^{4} 2 x^{3} d x)

\int_{1}^{4} 4 x d x = 30

\int_{1}^{4} 3 x^{2} d x = 63

\int_{1}^{4} 2 x^{3} d x = \frac{255}{2}

= - (30 + 63 - \frac{255}{2})

簡素化

- (30 + 63 - \frac{255}{2}) : \frac{69}{2}

= \frac{69}{2}