integral from 0 to 1 of |2x^2-1|

解答

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

2 \cdot \frac{2}{3} - \frac{1}{3}

十进制

0.60947 \dots

求解步骤

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

拿掉绝对值

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

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

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

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

化简

= 2 \cdot \frac{2}{3} - \frac{1}{3}

\frac{d x}{d t} = 2 x (5 - x)

y^{^{'}} = e^{2 y + 6 x}

\int \frac{8 x ^{3} + 8 x ^{2} + 8}{x ^{2} + x} d x

\int \frac{x ln ( x )}{x} d x

t an g e n t f (x) = \frac{1}{x ^{5}}, (2, \frac{1}{32})