integral from 1 to 6 of (x^3+x^2+3)

解答

\int_{1}^{6} (x^{3} + x^{2} + 3) d x

\frac{4925}{12}

十进制

410.41666 \dots

求解步骤

\int_{1}^{6} x^{3} + x^{2} + 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

= \int_{1}^{6} x^{3} d x + \int_{1}^{6} x^{2} d x + \int_{1}^{6} 3 d x

\int_{1}^{6} x^{3} d x = \frac{1295}{4}

\int_{1}^{6} x^{2} d x = \frac{215}{3}

\int_{1}^{6} 3 d x = 15

= \frac{1295}{4} + \frac{215}{3} + 15

化简

\frac{1295}{4} + \frac{215}{3} + 15 : \frac{4925}{12}

= \frac{4925}{12}