integral from 0 to 1 of 12x^2(1-x)

解答

\int_{0}^{1} 12 x^{2} (1 - x) d x

1

求解步骤

\int_{0}^{1} 12 x^{2} (1 - x) d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int_{0}^{1} x^{2} (1 - x) d x

乘开

x^{2} (1 - x) : x^{2} - x^{3}

= 12 \cdot \int_{0}^{1} x^{2} - x^{3} 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

= 12 (\int_{0}^{1} x^{2} d x - \int_{0}^{1} x^{3} d x)

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

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

= 12 (\frac{1}{3} - \frac{1}{4})

化简

12 (\frac{1}{3} - \frac{1}{4}) : 1

= 1

\int_{0}^{3} x^{2} - 4 x + 3 d x

\int_{5}^{25} \frac{1}{x ( ln ( x ) ) ^{9}} d x

\int_{0}^{5} \frac{1}{1 - x ^{2}} d x

\int_{0}^{\frac{1}{2}} x arctan (2 x) d x

\int_{2}^{5} ln (x) + sin (x) + 8 d x