integral from a to b of [3f(t)-2g(t)]

解

\int_{a}^{b} [3 f (t) - 2 g (t)] d t

- g (b^{2} - a^{2}) + \frac{3 f ( b ^{2} - a ^{2} )}{2}

解答ステップ

\int_{a}^{b} 3 f t - 2 g t 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_{a}^{b} 2 g t d t + \int_{a}^{b} 3 f t d t

\int_{a}^{b} 2 g t d t = g (b^{2} - a^{2})

\int_{a}^{b} 3 f t d t = \frac{3 f ( b ^{2} - a ^{2} )}{2}

= - g (b^{2} - a^{2}) + \frac{3 f ( b ^{2} - a ^{2} )}{2}