integral from 1 to 4 of 3x^2+xsqrt(x)+1

解答

\int_{1}^{4} 3 x^{2} + x x + 1 d x

\frac{392}{5}

十进制

78.4

求解步骤

\int_{1}^{4} 3 x^{2} + x 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_{1}^{4} 3 x^{2} d x + \int_{1}^{4} x x d x + \int_{1}^{4} 1 d x

\int_{1}^{4} 3 x^{2} d x = 63

\int_{1}^{4} x x d x = \frac{62}{5}

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

= 63 + \frac{62}{5} + 3

化简

63 + \frac{62}{5} + 3 : \frac{392}{5}

= \frac{392}{5}

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