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

解答

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

\frac{4}{3}

十进制

1.33333 \dots

求解步骤

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

乘开

(x - 1) (2 x + 1) : 2 x^{2} - x - 1

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

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

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

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

= \frac{16}{3} - 2 - 2

化简

\frac{16}{3} - 2 - 2 : \frac{4}{3}

= \frac{4}{3}

\int_{\frac{π}{2}}^{0} \frac{1 + cos ( 2 x )}{2} d x

\int_{- 1}^{0} 6 x^{2} (2 x^{3} + 2)^{2} d x

\int_{0}^{1} ln (r^{2} + 1) r d r

\int_{1}^{2} x^{3} + \frac{6}{x ^{2}} d x

\int_{2}^{4} \frac{1}{x ( ln ( x ) ) ^{9}} d x