integral from 4 to 9 of 2sqrt(t)(10t+3)

解答

\int_{4}^{9} 2 t (10 t + 3) d t

1764

求解步骤

\int_{4}^{9} 2 t (10 t + 3) d t

提出常数:

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

= 2 \cdot \int_{4}^{9} t (10 t + 3) d t

乘开

t (10 t + 3) : 10 t^{\frac{3}{2}} + 3 t

= 2 \cdot \int_{4}^{9} 10 t^{\frac{3}{2}} + 3 t d t

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

= 2 (\int_{4}^{9} 10 t^{\frac{3}{2}} d t + \int_{4}^{9} 3 t d t)

\int_{4}^{9} 10 t^{\frac{3}{2}} d t = 844

\int_{4}^{9} 3 t d t = 38

= 2 (844 + 38)

化简

= 1764

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