integral from 1 to x of (1-t)

解答

\int_{1}^{x} (1 - t) d t

x - 1 - \frac{x ^{2} - 1}{2}

求解步骤

\int_{1}^{x} 1 - 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

= \int_{1}^{x} 1 d t - \int_{1}^{x} t d t

\int_{1}^{x} 1 d t = x - 1

\int_{1}^{x} t d t = \frac{x ^{2} - 1}{2}

= x - 1 - \frac{x ^{2} - 1}{2}

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