integral from 0 to 1 of sqrt(x)(x+1)

解答

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

\frac{16}{15}

十进制

1.06666 \dots

求解步骤

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

乘开

x (x + 1) : x^{\frac{3}{2}} + x

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

\int_{0}^{1} x^{\frac{3}{2}} d x = \frac{2}{5}

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

= \frac{2}{5} + \frac{2}{3}

化简

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

\int \frac{e ^{0}}{( sin ( 3 x ) ) ^{2}} d x

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