integral from-1 to 1 of x(2x-sin(x^2))

解

\int_{- 1}^{1} x (2 x - sin (x^{2})) d x

\frac{4}{3}

十進法表記

1.33333 \dots

解答ステップ

\int_{- 1}^{1} x (2 x - sin (x^{2})) d x

拡張

x (2 x - sin (x^{2})) : 2 x^{2} - x sin (x^{2})

= \int_{- 1}^{1} 2 x^{2} - x sin (x^{2}) 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}^{1} 2 x^{2} d x - \int_{- 1}^{1} x sin (x^{2}) d x

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

\int_{- 1}^{1} x sin (x^{2}) d x = 0

= \frac{4}{3} - 0

簡素化

= \frac{4}{3}