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

解答

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

2 \cdot \frac{4 2 + 4}{15}

十进制

1.28758 \dots

求解步骤

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

提出常数:

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

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

使用换元积分法

= 2 \cdot \int_{1}^{2} (u - 1) u d u

乘开

(u - 1) u : u^{\frac{3}{2}} - u

= 2 \cdot \int_{1}^{2} u^{\frac{3}{2}} - u d u

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

= 2 (\int_{1}^{2} u^{\frac{3}{2}} d u - \int_{1}^{2} u d u)

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

\int_{1}^{2} u d u = \frac{4 2 - 2}{3}

= 2 (\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3})

化简

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3} : \frac{4 2 + 4}{15}

= 2 \cdot \frac{4 2 + 4}{15}

\int_{1}^{2} 1 + 64 x^{2} d x

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

\int_{t - 2}^{6} x - 4 d x

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

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