integral from-1 to 2 of sqrt(x+2)x^2

解答

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

\frac{562}{105}

十进制

5.35238 \dots

求解步骤

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

使用换元积分法

= \int_{1}^{4} (u - 2)^{2} u d u

乘开

(u - 2)^{2} u : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

= \int_{1}^{4} u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 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

= \int_{1}^{4} u^{\frac{5}{2}} d u - \int_{1}^{4} 4 u^{\frac{3}{2}} d u + \int_{1}^{4} 4 u d u

\int_{1}^{4} u^{\frac{5}{2}} d u = \frac{254}{7}

\int_{1}^{4} 4 u^{\frac{3}{2}} d u = \frac{248}{5}

\int_{1}^{4} 4 u d u = \frac{56}{3}

= \frac{254}{7} - \frac{248}{5} + \frac{56}{3}

化简

\frac{254}{7} - \frac{248}{5} + \frac{56}{3} : \frac{562}{105}

= \frac{562}{105}

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