integral of 18(200+t)^2

解答

\int 18 (200 + t)^{2} d t

18 (40000 t + 200 t^{2} + \frac{t ^{3}}{3}) + C

求解步骤

\int 18 (200 + t)^{2} d t

提出常数:

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

= 18 \cdot \int (200 + t)^{2} d t

乘开

(200 + t)^{2} : 40000 + 400 t + t^{2}

= 18 \cdot \int 40000 + 400 t + t^{2} 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

= 18 (\int 40000 d t + \int 400 t d t + \int t^{2} d t)

\int 40000 d t = 40000 t

\int 400 t d t = 200 t^{2}

\int t^{2} d t = \frac{t ^{3}}{3}

= 18 (40000 t + 200 t^{2} + \frac{t ^{3}}{3})

解答补常数

= 18 (40000 t + 200 t^{2} + \frac{t ^{3}}{3}) + C

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