integral from 2 to x of (t^3+8t-5)

解

\int_{2}^{x} (t^{3} + 8 t - 5) d t

4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10

解答ステップ

\int_{2}^{x} t^{3} + 8 t - 5 d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{x} t^{3} d t + \int_{2}^{x} 8 t d t - \int_{2}^{x} 5 d t

\int_{2}^{x} t^{3} d t = \frac{x ^{4}}{4} - 4

\int_{2}^{x} 8 t d t = 8 (\frac{x ^{2}}{2} - 2)

\int_{2}^{x} 5 d t = 5 x - 10

= \frac{x ^{4}}{4} - 4 + 8 (\frac{x ^{2}}{2} - 2) - (5 x - 10)

簡素化

\frac{x ^{4}}{4} - 4 + 8 (\frac{x ^{2}}{2} - 2) - (5 x - 10) : 4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10

= 4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10