integral from 1 to 0 of (3x^2+x-5)

解

\int_{1}^{0} (3 x^{2} + x - 5) d x

\frac{7}{2}

十進法表記

3.5

解答ステップ

\int_{1}^{0} 3 x^{2} + x - 5 d x

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

= - \int_{0}^{1} 3 x^{2} + x - 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_{0}^{1} 3 x^{2} d x + \int_{0}^{1} x d x - \int_{0}^{1} 5 d x)

\int_{0}^{1} 3 x^{2} d x = 1

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

\int_{0}^{1} 5 d x = 5

= - (1 + \frac{1}{2} - 5)

簡素化

- (1 + \frac{1}{2} - 5) : \frac{7}{2}

= \frac{7}{2}