integral from 1 to 4 of (x+2-x^2)

解答

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

- \frac{15}{2}

十进制

- 7.5

求解步骤

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

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} 2 d x = 6

\int_{1}^{4} x^{2} d x = 21

= \frac{15}{2} + 6 - 21

化简

\frac{15}{2} + 6 - 21 : - \frac{15}{2}

= - \frac{15}{2}

\int_{- 1}^{0} (x + 1)^{10} x d x

\int_{3}^{7} (8 x + 5) d x

\int_{a}^{7 a} \frac{1}{x} d x

\int_{0}^{u} \frac{t}{t + 1} d t

\int_{1}^{c o s (x)} 8 t^{7} d t