integral from 1 to 4 of (t+1/t)^2

解

\int_{1}^{4} (t + \frac{1}{t})^{2} d t

\frac{111}{4}

十進法表記

27.75

解答ステップ

\int_{1}^{4} (t + \frac{1}{t})^{2} d t

拡張

(t + \frac{1}{t})^{2} : t^{2} + 2 + \frac{1}{t ^{2}}

= \int_{1}^{4} t^{2} + 2 + \frac{1}{t ^{2}} 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_{1}^{4} t^{2} d t + \int_{1}^{4} 2 d t + \int_{1}^{4} \frac{1}{t ^{2}} d t

\int_{1}^{4} t^{2} d t = 21

\int_{1}^{4} 2 d t = 6

\int_{1}^{4} \frac{1}{t ^{2}} d t = \frac{3}{4}

= 21 + 6 + \frac{3}{4}

簡素化

21 + 6 + \frac{3}{4} : \frac{111}{4}

= \frac{111}{4}