integral from 3 to 5 of 6x^2(x^3+4)^3

解答

\int_{3}^{5} 6 x^{2} (x^{3} + 4)^{3} d x

137999680

求解步骤

\int_{3}^{5} 6 x^{2} (x^{3} + 4)^{3} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{3}^{5} x^{2} (x^{3} + 4)^{3} d x

乘开

x^{2} (x^{3} + 4)^{3} : x^{11} + 12 x^{8} + 48 x^{5} + 64 x^{2}

= 6 \cdot \int_{3}^{5} x^{11} + 12 x^{8} + 48 x^{5} + 64 x^{2} 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

= 6 (\int_{3}^{5} x^{11} d x + \int_{3}^{5} 12 x^{8} d x + \int_{3}^{5} 48 x^{5} d x + \int_{3}^{5} 64 x^{2} d x)

\int_{3}^{5} x^{11} d x = \frac{60902296}{3}

\int_{3}^{5} 12 x^{8} d x = \frac{7733768}{3}

\int_{3}^{5} 48 x^{5} d x = 119168

\int_{3}^{5} 64 x^{2} d x = \frac{6272}{3}

= 6 (\frac{60902296}{3} + \frac{7733768}{3} + 119168 + \frac{6272}{3})

化简

6 (\frac{60902296}{3} + \frac{7733768}{3} + 119168 + \frac{6272}{3}) : 137999680

= 137999680

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