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

解答

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

- \frac{7}{12}

十进制

- 0.58333 \dots

求解步骤

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

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

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

= \frac{1}{4} + \frac{2}{3} - \frac{3}{2}

化简

\frac{1}{4} + \frac{2}{3} - \frac{3}{2} : - \frac{7}{12}

= - \frac{7}{12}

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

\int_{0}^{25} 85 - 3 25 - x d x

\int_{- 2}^{2} t^{3} d t

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{4}} d x

\int_{0}^{1} (x^{4}) d x