integral from-2 to 2 of x^2+ax^3+bx^4

解

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

\frac{64 b}{5} + \frac{16}{3}

解答ステップ

\int_{- 2}^{2} x^{2} + a x^{3} + b x^{4} 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_{- 2}^{2} x^{2} d x + \int_{- 2}^{2} a x^{3} d x + \int_{- 2}^{2} b x^{4} d x

\int_{- 2}^{2} x^{2} d x = \frac{16}{3}

\int_{- 2}^{2} a x^{3} d x = 0

\int_{- 2}^{2} b x^{4} d x = \frac{64 b}{5}

= \frac{16}{3} + 0 + \frac{64 b}{5}

簡素化

= \frac{64 b}{5} + \frac{16}{3}