integral from 0 to 1 of (4x+1)^2

解答

\int_{0}^{1} (4 x + 1)^{2} d x

\frac{31}{3}

十进制

10.33333 \dots

求解步骤

\int_{0}^{1} (4 x + 1)^{2} d x

乘开

(4 x + 1)^{2} : 16 x^{2} + 8 x + 1

= \int_{0}^{1} 16 x^{2} + 8 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}^{1} 16 x^{2} d x + \int_{0}^{1} 8 x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} 16 x^{2} d x = \frac{16}{3}

\int_{0}^{1} 8 x d x = 4

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

= \frac{16}{3} + 4 + 1

化简

\frac{16}{3} + 4 + 1 : \frac{31}{3}

= \frac{31}{3}

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